Question

given that z^1/2 = X^1/2 +Y^1/2, show that (x+y-z)^2 = 4xy

Answer 1

Answer:

   √z = √x + √y

⇒ z = (√x + √y)² = x + y + 2√(xy)

⇒ x + y - z = -2√(xy)

⇒ (x + y - z)² = 4xy

Answer 2

Answer:

(x+y-z)^2 = 4xy is verified.

Step-by-step explanation:

Given,

z^1/2 = X^1/2 +Y^1/2

$z^{1/2} = x^{1/2} + y^{1/2}\\ \\\sqrt{z} = \sqrt{x} + \sqrt{y} \\\\(\sqrt{z}) ^{2} = (\sqrt{x} + \sqrt{y} ) ^{2}\\\\z = (\sqrt{x}) ^{2} + (\sqrt{y}) ^{2} + 2\sqrt{xy} \\\\z = x + y + 2\sqrt{xy}$    (∵ (a+b)² =a²+b²+2ab )

We need to prove,  (x+y-z)^2 = 4xy

Consider L.H.S,

L.H.S = (x+y-z)^2

$( x+y-z)^{2}\\\\= (x + y - (x+y+2\sqrt{xy} ))^{2}\\\\= (x+y -x-y-2\sqrt{xy})^{2}\\ \\= (2\sqrt{xy}) ^{2}\\\\= 2^{2} * (\sqrt{xy})^{2} \\\\= 4 * xy\\\\= 4xy$        ( ∵ z = x + y + 2√xy )

= 4xy = R.H.S

∴ L.H.S = R.H.S

Hence, we showed that (x+y-z)^2 = 4xy

Click here for more about the formulas used:

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