Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
 
(22 intermediate revisions by 10 users not shown)
Line 2: Line 2:
 
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>?
  
−
== Solution ==
+
== Solution 1 ==
−
If we expand by squaring, we get a quartic [[polynomial]], which obviously isn't very helpful.
+
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.
  
−
Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our equation becomes <math>y=2\sqrt{y+15}</math>.
+
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. 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>,
−
<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>.
+
 
 +
<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 ==
 +
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>.
 +
 
 +
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>
 +
 
 +
Reasoning as in Solution 1, the product of the roots is <math>\boxed{020}</math>.
 +
 
 +
== Solution 3 ==
 +
 
 +
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>
 +
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>
 +
The solutions in <math>y</math> are then <cmath>y=x+9=\pm3\sqrt{5},\pm\sqrt{61}</cmath>
 +
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>
 +
By difference of squares.
 +
 
 +
== Solution 4 ==
 +
 
 +
We are given the equation
 +
<cmath>x^2+18x+30=2\sqrt{x^2+18x+45}</cmath>
 +
Squaring both sides yields
 +
<cmath>(x^2+18x+30)^2=4(x^2+18x+45)</cmath>
 +
<cmath>(x^2+18x+30)^2=4(x^2+18x+30+15)</cmath>
 +
<cmath>(x^2+18x+30)^2=4(x^2+18x+30)+60</cmath>
 +
<cmath>(x^2+18x+30)^2-4(x^2+18x+30)-60=0</cmath>
 +
Substituting <math>y=x^2+18x+30</math> yields
 +
<cmath>y^2-4y-60=0</cmath>
 +
<cmath>(y+6)(y-10)=0</cmath>
 +
Thus <math>y=x^2+18x+30=-6,10</math>. However if <math>y=-6</math>, the left side of the equation
 +
<cmath>x^2+18x+30=2\sqrt{x^2+18x+45}</cmath>
 +
would be negative while the right side is negative. Thus <math>y=10</math> is the only possible value and we have
 +
<cmath>x^2+18x+30=10</cmath>
 +
<cmath>x^2+18x+20=0</cmath>
 +
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>.
 +
 
 +
~ Nafer
  
 
== See Also ==
 
== See Also ==

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_3&oldid=176109"