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| ~MRENTHUSIASM | | ~MRENTHUSIASM |
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| ==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> |
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| so the answer is <math>141+4=\boxed{145}.</math> | | so the answer is <math>141+4=\boxed{145}.</math> |
| ~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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− |
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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== | | ==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 | | ~hihitherethere |
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