Question

For how many natural numbers a is the expression $\sqrt{\frac{a+64}{a-64} }$ also a natural number?

Answer 1

Required Answer.

Only one value.

Before we solve.

The square root should be a natural number, so the number is a perfect square.

Solution.

Let $\sqrt{\dfrac{a+64}{a-64}} =k$ where $k$ is a natural number.

Then $\dfrac{a+64}{a-64} =k^2$.

$\rightarrow \dfrac{(a-64)+64+64}{a-64} =k^2$

$\rightarrow 1+\dfrac{128}{a-64} =k^2$

$\therefore k^2-1=\dfrac{128}{a-64}$

However, $\dfrac{128}{a-64}$ can only give a factor of $128$. Hence, possible values of the LHS are $k^2-1=1,2,4,8,16,32,64,128$.

Hence the possible value of $k^2$ is $9$ only.

There is only 1 value of a, which is $a=80$, such that $\sqrt{\dfrac{a+64}{a-64} } =3$.

Answer 2

Answer:

Solution :-

Let us assume

$\bf \: \sqrt{ \dfrac{a + 64}{a - 64} } = x$

On squaring both sides

$\bf \sqrt{ {\dfrac{ {a + 64} }{a - 64} }^{2} } = {x}^{2}$

$\bf \dfrac{a + 64}{a - 64} = {x}^{2}$

$\sf \dfrac{(a - 64) + 64 + 64}{a - 64} = x {}^{2}$

$\sf \: \dfrac{ \cancel {a - 64} + 128}{ \cancel{a - 64}} = {x}^{2}$

$\sf \: 1 + \dfrac{128}{a - 64} = x {}^{2}$

$\sf \: \dfrac{128}{a - 64} = {x }^{2} - 1$

Possible value is 80

• V E R I F I C A T I O N

$\bf \: \sqrt{ \dfrac{80 + 64}{80 - 64} } = 3$

$\bf \sqrt{ \dfrac{144}{16} } = 3$

$\sqrt{9} = 3$

$3 = 3$