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1963 AHSME Problems/Problem 40

Problem

If $x$ is a number satisfying the equation $\sqrt{x+9}-\sqrt{x-9}=3$, 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 1

Let $a = \sqrt{x + 9}$ and $b = \sqrt{x - 9}$. 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 $\sqrt{x + 9} - \sqrt{x - 9} = 3$, 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 $a = \sqrt{x + 9}$ and $b = \sqrt{x - 9}$, so $ab = \sqrt{(x + 9)(x - 9)} = \sqrt{x^2 - 81}$, so $\sqrt{x^2 - 81} = -1$. Cubing both sides, we get $x^2 - 81 = -1$, so $x^2 = 80$. The answer is $\boxed{\textbf{(C)}}$.

Solution 2 $\sqrt{x+9}-\sqrt{x-9}-3=0$ i.e, $\sqrt{x+9}+\sqrt{-x+9}+(-3)=0$

if the sum of three numbers is zero, then their sum of cubes is thrice the product of each number.
then, $x+9-x+9+27=3(\sqrt{x+9})(\sqrt{9-x})(-3)$ . by solving this, we get $-9=-9\sqrt{9^2-x^2}$ $x^2=80$. this gives the step what we had done in solution 1.The answer is $\boxed{\textbf{(C)}}$.

Solution 3

Cubing both sides to get rid of the messy cube roots, we get $$x+9-3\sqrt{(x+9)^2(x-9)}+3\sqrt{(x+9)(x-9)^2}-x+9=27,$$so $$18-3\sqrt{(x-9)(x+9)}(\sqrt{x+9}-\sqrt{x-9})=27\implies 3\sqrt{x^2-81}=-3.$$ Dividing both sides by 3 and cubing, we find $x^2=80$, which is between $\boxed{(C)75\text{ and }85}$.

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