Question

If x and y are two positive real numbers such that x 2 +4y 2 =17 and xy =2, then find the value of x- 2y. a. 3 b. 4 c. 8 d. 9

Answer 1

Answer: The value of x- 2y is a.
`\pm 3`.

Explanation:

Given: x and y are two positive real numbers such that
`x^2+4y^2=17` and
`xy= 2` .

Consider
`(x-2y)^2=x^2-2(x)(2y)+(2y)^2\ \ \ [(a+b)^2=(a^2-2ab+b^2)]`

`=x^2-4xy+4y^2`

`=x^2+4y^2-4(xy)`

Put
`x^2+4y^2=17` and
`xy= 2` , we get

`(x-2y)^2=17-4(2)=17-8=9`

`\Rightarrow\ (x-2y)^2=9`

Taking square root on both sides , we get'

`x-2y= \pm3`

Hence, the value of x- 2y is a.
`\pm 3`.