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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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== Solutions ==
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=== Solution 1 ===
 
If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with.
 
If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with.
  
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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 the roots is <math>\boxed{020}</math>.
 
<center><math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.</math></center> By [[Vieta's formulas]], the product of the roots is <math>\boxed{020}</math>.
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=== 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+ 18 + 45) - 2\sqrt{x^2+18+45} - 15 = 0.</cmath> Letting <math>n = \sqrt{x^2+18+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+18+45} = 5 \Longrightarrow x^2 + 18 + 45 = 25 \Longrightarrow x^2+18+20=0.</cmath>
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And by [[Vieta's formulas]], the product of the roots is <math>\boxed{020}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 11:25, 15 October 2016

Problem

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

Solutions

Solution 1

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

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

Now we can square; solving for $y$, we get $y=10$ or $y=-6$ (The second solution is extraneous since $2\sqrt{y+15}$ is positive (plugging in $6$ as $y$, we get $-$$6$ $=$ $6$, which is obviously not true)).So, we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

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

By Vieta's formulas, the product of the roots is $\boxed{020}$.

Solution 2

We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \[(x^2+ 18 + 45) - 2\sqrt{x^2+18+45} - 15 = 0.\] Letting $n = \sqrt{x^2+18+45}$, 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 \[\sqrt{x^2+18+45} = 5 \Longrightarrow x^2 + 18 + 45 = 25 \Longrightarrow x^2+18+20=0.\]

And by Vieta's formulas, the product of the roots is $\boxed{020}$.

See Also

1983 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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