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Difference between revisions of "1963 AHSME Problems/Problem 40"

(Created page with "Let <math>a = \sqrt[3]{x + 9}</math> and <math>b = \sqrt[3]{x - 9}</math>. Cubing these equations, we get <math>a^3 = x + 9</math> and <math>b^3 = x - 9</math>, so <math>a^3 -...")
 
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== Problem 40==
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If <math>x</math> is a number satisfying the equation <math>\sqrt[3]{x+9}-\sqrt[3]{x-9}=3</math>, then <math>x^2</math> is between:
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<math>\textbf{(A)}\ 55\text{ and }65\qquad
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\textbf{(B)}\ 65\text{ and }75\qquad
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\textbf{(C)}\ 75\text{ and }85\qquad
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\textbf{(D)}\ 85\text{ and }95\qquad
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\textbf{(E)}\ 95\text{ and }105 </math>
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==Solution==
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Let <math>a = \sqrt[3]{x + 9}</math> and <math>b = \sqrt[3]{x - 9}</math>. Cubing these equations, we get <math>a^3 = x + 9</math> and <math>b^3 = x - 9</math>, so <math>a^3 - b^3 = 18</math>. The left-hand side factors as
 
Let <math>a = \sqrt[3]{x + 9}</math> and <math>b = \sqrt[3]{x - 9}</math>. Cubing these equations, we get <math>a^3 = x + 9</math> and <math>b^3 = x - 9</math>, so <math>a^3 - b^3 = 18</math>. The left-hand side factors as
 
<cmath>(a - b)(a^2 + ab + b^2) = 18.</cmath>
 
<cmath>(a - b)(a^2 + ab + b^2) = 18.</cmath>
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However, from the given equation <math>\sqrt[3]{x + 9} - \sqrt[3]{x - 9} = 3</math>, we get <math>a - b = 3</math>. Then <math>3(a^2 + ab + b^2) = 18</math>, so <math>a^2 + ab + b^2 = 18/3 = 6</math>.
 
However, from the given equation <math>\sqrt[3]{x + 9} - \sqrt[3]{x - 9} = 3</math>, we get <math>a - b = 3</math>. Then <math>3(a^2 + ab + b^2) = 18</math>, so <math>a^2 + ab + b^2 = 18/3 = 6</math>.
  
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Squaring the equation <math>a - b = 3</math>, we get <math>a^2 - 2ab + b^2 = 9</math>. Subtracting this equation from the equation <math>a^2 + ab + b^2 = 6</math>, we get <math>3ab = -3</math>, so <math>ab = -1</math>. But <math>a = \sqrt[3]{x + 9}</math> and <math>b = \sqrt[3]{x - 9}</math>, so <math>ab = \sqrt[3]{(x + 9)(x - 9)} = \sqrt[3]{x^2 - 81}</math>, so <math>\sqrt[3]{x^2 - 81} = -1</math>. Cubing both sides, we get <math>x^2 - 81 = -1</math>, so <math>x^2 = \boxed{80}</math>. The answer is (C).
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Squaring the equation <math>a - b = 3</math>, we get <math>a^2 - 2ab + b^2 = 9</math>. Subtracting this equation from the equation <math>a^2 + ab + b^2 = 6</math>, we get <math>3ab = -3</math>, so <math>ab = -1</math>. But <math>a = \sqrt[3]{x + 9}</math> and <math>b = \sqrt[3]{x - 9}</math>, so <math>ab = \sqrt[3]{(x + 9)(x - 9)} = \sqrt[3]{x^2 - 81}</math>, so <math>\sqrt[3]{x^2 - 81} = -1</math>. Cubing both sides, we get <math>x^2 - 81 = -1</math>, so <math>x^2 = 80</math>. The answer is <math>\boxed{\textbf{(C)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1963|num-b=39|after=Last Problem}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 01:59, 7 June 2018

Problem 40

If $x$ is a number satisfying the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^2$ is between:

$\textbf{(A)}\ 55\text{ and }65\qquad \textbf{(B)}\ 65\text{ and }75\qquad \textbf{(C)}\ 75\text{ and }85\qquad \textbf{(D)}\ 85\text{ and }95\qquad \textbf{(E)}\ 95\text{ and }105$

Solution

Let $a = \sqrt[3]{x + 9}$ and $b = \sqrt[3]{x - 9}$. Cubing these equations, we get $a^3 = x + 9$ and $b^3 = x - 9$, so $a^3 - b^3 = 18$. The left-hand side factors as \[(a - b)(a^2 + ab + b^2) = 18.\]

However, from the given equation $\sqrt[3]{x + 9} - \sqrt[3]{x - 9} = 3$, we get $a - b = 3$. Then $3(a^2 + ab + b^2) = 18$, so $a^2 + ab + b^2 = 18/3 = 6$.

Squaring the equation $a - b = 3$, we get $a^2 - 2ab + b^2 = 9$. Subtracting this equation from the equation $a^2 + ab + b^2 = 6$, we get $3ab = -3$, so $ab = -1$. But $a = \sqrt[3]{x + 9}$ and $b = \sqrt[3]{x - 9}$, so $ab = \sqrt[3]{(x + 9)(x - 9)} = \sqrt[3]{x^2 - 81}$, so $\sqrt[3]{x^2 - 81} = -1$. Cubing both sides, we get $x^2 - 81 = -1$, so $x^2 = 80$. The answer is $\boxed{\textbf{(C)}}$.

See Also

1963 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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