Radical problem

Algebra Level 2

$\large \sqrt[x]{64} \cdot \sqrt[y]{64} = 32$

Let $x$ and $y$ be positive integers satisfying the equation above. Find $x+y$ .

The answer is 5.

2 solutions

Tapas Mazumdar
Dec 23, 2016

$\begin{aligned} & \sqrt[x]{64} \cdot \sqrt[y]{64} = 32 \\ \implies & 2^{{6}/{x}} \cdot 2^{{6}/{y}} = 2^5 \\ \implies & 2^{6 \left( \frac 1x + \frac 1y \right)} = 2^5 \\ \implies & 6 \left( \dfrac 1x + \dfrac 1y \right) = 5 \\ \implies & \dfrac 1x + \dfrac 1y = \dfrac 56 \\ \implies & \dfrac 1x + \dfrac 1y = \dfrac 12 + \dfrac 13 \\ \implies & (x,y) = (2,3) ; (3,2) \\ \implies & x+y = \boxed{5} \end{aligned}$

Splendidi , you figured it out :)

Raffaele Piccirillo - 4 years, 5 months ago

Thank you!

Tapas Mazumdar - 4 years, 5 months ago

Raffaele Piccirillo
Dec 23, 2016

This is it

$\dfrac{x+y}{xy} = \dfrac 56$ $\implies x+y = 5 \ \ \text{and} \ \ xy=6$

That doesn't seem right to me. In particular if $\dfrac ab = \dfrac cd$ , would it be necessarily true that $a=c$ and $b=d$ ?

Actually $(2,3)$ or $(3,2)$ are the only integral pairs that would satisfy this equation and thus your assumption was correct because these two integers are co-prime to each other.

But for instance take $\dfrac 1x + \dfrac 1y = \dfrac 49$ then by your assumption,

$x+y = 4 \\ xy = 9$

whereas, the actual ordered pairs for this equation for positive integers $x$ and $y$ is $(9,3)$ and $(3,9)$ .

Thus, $x+y = 12$ not $4$ . Also, $xy=27$ not $9$ .

Please take care of such small misconceptions in the future.

Tapas Mazumdar - 4 years, 5 months ago

I already solved the second degree equation for the system[x+y=5;xy=6] , so I put 5 as the only result :)

Raffaele Piccirillo - 4 years, 5 months ago

Radical problem

The answer is 5.

2 solutions

0 pending reports