Question

Find two numbers such that the sum of their squares is 100. The square of the larger number is equal to the sum of the product of the larger numberand four times the smaller number and eight times the larger number.

Answer 1

Answer with Step-by-step explanation:

Let x and y are two  numbers

According to question

$x^2+y^2=100$....(1)

Suppose x is larger than y

$x^2=x(4y+8x)$

$x=4y+8x$

$x-8x=4y$

$-7x=4y$

$y=-\frac{7}{4}x$

Substitute the value of y

$x^2+(-\frac{7}{4}x)^2=100$

$x^2+\frac{49}{16}x^2=100$

$\frac{16x^2+49x^2}{16}=100$

$\frac{65}{16}x^2=100$

$x^2=\frac{16\times 100}{65}$

$x=\sqrt{\frac{1600}{65}}=\frac{40}{\sqrt{65}}=\frac{40}{65}\sqrt{65}=\frac{8}{13}\sqrt{65}$

$y=-\frac{7}{4}\times \frac{8}{13}\sqrt{65}=-\frac{14}{13}\sqrt{65}$