Question

The square root of x2/y2 + y2/4x2 - x/y + y/2x - 3/4
Plz say with explanation

Answer 1

Answer:

(2x/y + 1)(2x^3/y^3 - 3x^2/y^2 + 1) ] / 2x/y

Step-by-step explanation:

Let Q be the given expression. Putting x/y = a in Q we get,

Q = a^2 + 1/(4 *a^2) - a + 1/2a - 3/4

= (a^2 -a - 3/4) + (1/2a + 1/4a^2) = (4a^2 - 4a - 3)/4 + (2a+1)/4a^2

= (2a+1)(2a-3)/4 + (2a+1)/4a^2 = (2a+1)/4 * { (2a-3) + 1/a^2}

=(2a+1)/4 * (2a^3 - 3a^2+ 1)/a^2 = (2a+1)(2a^3–3a^2+1)/4a^2

So, SQRT(Q) = [sqrt { (2a+1)(2a^3–3a^2+1)}]/2a. Replacing the value of a as x/y, we get the desired squared root.

= SQRT [ (2x/y + 1)(2x^3/y^3 - 3x^2/y^2 + 1) ] / 2x/y