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. $\pm 3$.

Step-by-step 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. $\pm 3$.