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| ~MRENTHUSIASM | | ~MRENTHUSIASM |
− | ==Solution 4 (way too long)== | + | |
| + | ==Solution 4 (Way Too Long)== |
| Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: <cmath>abc+d(ab+bc+ca)=14</cmath> | | Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: <cmath>abc+d(ab+bc+ca)=14</cmath> |
| Then we plug in equation 2 to receive <math>abc-4d=14</math>. By equation 4 we get <math>abc=\frac{30}{d}</math>. Plugging in, we get <math>\frac{30}{d}-4d=14</math>. Multiply by <math>d</math> on both sides to get the quadratic equation <math>4d^2+14d-30=0</math>. Solving using the quadratic equation, we receive <math>d=\frac{3}{2},d=-5</math>. So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get <cmath>ab+c(a+b)=-4</cmath> | | Then we plug in equation 2 to receive <math>abc-4d=14</math>. By equation 4 we get <math>abc=\frac{30}{d}</math>. Plugging in, we get <math>\frac{30}{d}-4d=14</math>. Multiply by <math>d</math> on both sides to get the quadratic equation <math>4d^2+14d-30=0</math>. Solving using the quadratic equation, we receive <math>d=\frac{3}{2},d=-5</math>. So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get <cmath>ab+c(a+b)=-4</cmath> |
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| Let <math>c=\frac{10}{3}</math>. | | Let <math>c=\frac{10}{3}</math>. |
| Then <math>ab=6</math>, so <math>a=\frac{6}{b}</math>. We thus get <math>\frac{6}{b}+b=-3</math>, which leads to the quadratic <math>b^2+3b+6</math>. The discriminant for this is <math>9-24</math>. That means this value of <math>c</math> is wrong, so <math>c=-2</math>. Thus we get polynomial <math>b^2+3b-10</math>. The discriminant this time is <math>49</math>, so we get two values for <math>b</math>. Through simple inspection, you may see they are interchangeable, as if you take the value <math>b=2</math>, you get <math>a=-5</math>. If you take the value <math>b=-5</math>, you get <math>a=2</math>. So it doesn't matter. That means the sum of all their squares is | | Then <math>ab=6</math>, so <math>a=\frac{6}{b}</math>. We thus get <math>\frac{6}{b}+b=-3</math>, which leads to the quadratic <math>b^2+3b+6</math>. The discriminant for this is <math>9-24</math>. That means this value of <math>c</math> is wrong, so <math>c=-2</math>. Thus we get polynomial <math>b^2+3b-10</math>. The discriminant this time is <math>49</math>, so we get two values for <math>b</math>. Through simple inspection, you may see they are interchangeable, as if you take the value <math>b=2</math>, you get <math>a=-5</math>. If you take the value <math>b=-5</math>, you get <math>a=2</math>. So it doesn't matter. That means the sum of all their squares is |
− | <cmath>\frac{9}{4}+4+4+25=\frac{141}{4}</cmath> | + | <cmath>\frac{9}{4}+4+4+25=\frac{141}{4},</cmath> |
| so the answer is <math>141+4=\boxed{145}.</math> | | so the answer is <math>141+4=\boxed{145}.</math> |
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| ~amcrunner | | ~amcrunner |
− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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| ==Solution 5== | | ==Solution 5== |
| Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>. | | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>. |
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math> | + | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>. |
− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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− | ~hihitherethere
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− | ==Solution 5==
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− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
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− | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>
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| ~hihitherethere | | ~hihitherethere |
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