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

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== Solutions ==
 
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with.
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If we were to expand by squaring, we would get a 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>.
  
−
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 (plugging in <math>6</math> as <math>y</math>, we get <math>-</math><math>6</math> <math>=</math> <math>6</math>, which is obviously not true)).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>,
+
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 the roots is <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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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>
 
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>
  
−
And by [[Vieta's formulas]], the product of the roots is <math>\boxed{020}</math>.
+
Reasoning as in Solution 1, the product of the roots is <math>\boxed{020}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 19:08, 15 February 2019

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 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 $\boxed{020}$.

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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