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Difference between revisions of "1983 AIME Problems/Problem 3"

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What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?
 
What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?
  
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== Solution ==
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== Solution 1 ==
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If we expand by squaring, we get a quartic [[polynomial]], which obviously isn't very helpful.
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If we were to expand by squaring, we would get a [[quartic Equation|quartic]] [[polynomial]], which isn't always the easiest thing to deal with.
  
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Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our equation becomes <math>y=2\sqrt{y+15}</math>.
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Instead, we substitute <math>y</math> for <math>x^2+18x+30</math>, so that the equation becomes <math>y=2\sqrt{y+15}</math>.
  
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Now we can square; solving for <math>y</math>, we get <math>y=10</math> or <math>y=-6</math>. The second solution is extraneous since <math>2\sqrt{y=15}</math> is positive. So, we have <math>y=10</math> as the only solution for <math>y</math>. Substituting <math>x^2+18x+30</math> back in for <math>y</math>,
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Now we can square; solving for <math>y</math>, we get <math>y=10</math> or <math>y=-6</math>. The second root is extraneous since <math>2\sqrt{y+15}</math> is always non-negative (and moreover, plugging in <math>y=-6</math>, we get <math>-6=6</math>, which is obviously false). Hence we have <math>y=10</math> as the only solution for <math>y</math>. Substituting <math>x^2+18x+30</math> back in for <math>y</math>,
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<center><math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.</math></center> By [[Vieta's formulas]], the product of our roots is therefore <math>\boxed{020}</math>.
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<center><math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.</math></center> Both of the roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's formulas]], the product of the real roots is simply <math>\boxed{020}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
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We begin by noticing that the polynomial on the left is <math>15</math> less than the polynomial under the radical sign. Thus: <cmath>(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.</cmath> Letting <math>n = \sqrt{x^2+18x+45}</math>, we have <math>n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0</math>. Because the square root of a real number can't be negative, the only possible <math>n</math> is <math>5</math>.
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Substituting that in, we have <cmath>\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.</cmath>
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Reasoning as in Solution 1, the product of the roots is <math>\boxed{020}</math>.
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== Solution 3 ==
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Begin by completing the square on both sides of the equation, which gives <cmath>(x+9)^2-51=2\sqrt{(x+3)(x+15)}</cmath>
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Now by substituting <math>y=x+9</math>, we get <math>y^2-51=2\sqrt{(y-6)(y+6)}</math>, or <cmath>y^4-106y^2+2745=0</cmath>
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The solutions in <math>y</math> are then <cmath>y=x+9=\pm3\sqrt{5},\pm\sqrt{61}</cmath>
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Turns out, <math>\pm3\sqrt{5}</math> are a pair of extraneous solutions. Thus, our answer is then <cmath>\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}</cmath>
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By difference of squares.
  
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There is an error with solution one since the second solution for <math>y</math> does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have:
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== Solution 4 ==
  
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<math>x^4+36x^3+380x^2+1080x+900 = 0</math>
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We are given the equation
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<cmath>x^2+18x+30=2\sqrt{x^2+18x+45}</cmath>
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Squaring both sides yields
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<cmath>(x^2+18x+30)^2=4(x^2+18x+45)</cmath>
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<cmath>(x^2+18x+30)^2=4(x^2+18x+30+15)</cmath>
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<cmath>(x^2+18x+30)^2=4(x^2+18x+30)+60</cmath>
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<cmath>(x^2+18x+30)^2-4(x^2+18x+30)-60=0</cmath>
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Substituting <math>y=x^2+18x+30</math> yields
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<cmath>y^2-4y-60=0</cmath>
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<cmath>(y+6)(y-10)=0</cmath>
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Thus <math>y=x^2+18x+30=-6,10</math>. However if <math>y=-6</math>, the left side of the equation
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<cmath>x^2+18x+30=2\sqrt{x^2+18x+45}</cmath>
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would be negative while the right side is negative. Thus <math>y=10</math> is the only possible value and we have
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<cmath>x^2+18x+30=10</cmath>
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<cmath>x^2+18x+20=0</cmath>
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Since the discriminant <math>\sqrt{18^2-4\cdot20}</math> is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, <math>\boxed{20}</math>.
  
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By Vieta's formula we get that the product is <math>\boxed{900}</math>.
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~ Nafer
  
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== See also ==
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== See Also ==
 
{{AIME box|year=1983|num-b=2|num-a=4}}
 
{{AIME box|year=1983|num-b=2|num-a=4}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 04:18, 21 July 2022

Problem

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?

Solution 1

If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with.

Instead, we substitute $y$ for $x^2+18x+30$, so that the equation becomes $y=2\sqrt{y+15}$.

Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$, we get $-6=6$, which is obviously false). Hence we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$

Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$, which is positive. Thus by Vieta's formulas, the product of the real roots is simply $\boxed{020}$.

Solution 2

We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \[(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.\] Letting $n = \sqrt{x^2+18x+45}$, we have $n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0$. Because the square root of a real number can't be negative, the only possible $n$ is $5$.

Substituting that in, we have \[\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.\]

Reasoning as in Solution 1, the product of the roots is $\boxed{020}$.

Solution 3

Begin by completing the square on both sides of the equation, which gives \[(x+9)^2-51=2\sqrt{(x+3)(x+15)}\] Now by substituting $y=x+9$, we get $y^2-51=2\sqrt{(y-6)(y+6)}$, or \[y^4-106y^2+2745=0\] The solutions in $y$ are then \[y=x+9=\pm3\sqrt{5},\pm\sqrt{61}\] Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \[\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}\] By difference of squares.

Solution 4

We are given the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] Squaring both sides yields \[(x^2+18x+30)^2=4(x^2+18x+45)\] \[(x^2+18x+30)^2=4(x^2+18x+30+15)\] \[(x^2+18x+30)^2=4(x^2+18x+30)+60\] \[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\] Substituting $y=x^2+18x+30$ yields \[y^2-4y-60=0\] \[(y+6)(y-10)=0\] Thus $y=x^2+18x+30=-6,10$. However if $y=-6$, the left side of the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \[x^2+18x+30=10\] \[x^2+18x+20=0\] Since the discriminant $\sqrt{18^2-4\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\boxed{20}$.

~ Nafer

See Also

1983 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions
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