Question

please batao koi please​

Answer 1

$\mathsf{Given :\;a^{\frac{1}{2}} + b^{\frac{1}{2}} - c^{\frac{1}{2}} = 0}$

$\mathsf{\implies a^{\frac{1}{2}} + b^{\frac{1}{2}} = c^{\frac{1}{2}}}$

Squaring on both sides, We get :

$\implies \mathsf{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right)^2 = \left(c^{\frac{1}{2}}\right)^2}$

Left side expression is in form of Identity : (P + Q)²

★  We know that : (P + Q)² = P² + Q² + 2PQ

$\implies \mathsf{\left(a^{\frac{1}{2}}\right)^2 + \left(b^{\frac{1}{2}}\right)^2 + 2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right) = \left(c^{\frac{1}{2}}\right)^2}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{\left(p^{n}\right)^m = p^{mn}}}}$

$\implies \mathsf{\left(a^{\frac{2}{2}}\right) + \left(b^{\frac{2}{2}}\right) + 2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right) = \left(c^{\frac{2}{2}}\right)}$

$\implies \mathsf{a + b + 2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right) = c}$

$\implies \mathsf{a + b - c = -2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right)}$

Squaring on both sides, We get :

$\implies \mathsf{(a + b - c)^2 = (-1)^2(2)^2\left(a^{\frac{1}{2}}\right)^2\left(b^{\frac{1}{2}}\right)^2}$

$\implies \mathsf{(a + b - c)^2 = 4\left(a^{\frac{2}{2}}\right)\left(b^{\frac{2}{2}}\right)}$

$\implies \mathsf{(a + b - c)^2 = 4ab}$

Answer : Option (3)

Answer 2

Solution :-

Given, √a + √b - √c = 0

       ⇒ √a + √b = √c            

( Squaring on both sides )

       ⇒ (√a + √b)² = √c²

       ⇒ a + b + 2√ab = c

       ⇒ (a + b - c) = -2 √ab

Again squaring on both sides, we get,

           (a + b - c)² = (-2 √ab)²

       ∴  (a + b - c)² = 4ab