Question

(√a+√b) (√a-√b) = a-b​

Answer 1

Answer:

On Squaring both sides of the equation

(√a+√b)² = (√(a+b))²

√a(√a+√b) + √b(√a+√b) = a + b

a + √ab + √ab + b = a + b

a + 2(√ab) + b ≠ a + b

Provided a ≠ 0, and b ≠ 0

LHS ≠ RHS

Let us assume a = 3, while b = 12

3 + 2(√36) + 12 ≠ 3 + 12

3 + 2(6) + 12 ≠ 3 + 12

3 + 12 + 12 ≠ 15

27 ≠ 15

Answer 2

Answer:

The expression in the L.H.S.  $(\sqrt{a} +\sqrt{b})(\sqrt{a}-\sqrt{b})$ when simplified gives the value  $(a-b)$ which is R.H.S. Hence it is proved.

Step-by-step explanation:

We have to prove an expression that is written as

$(\sqrt{a} +\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$

Let us break the expression as L.H.S. equation and R.H.S. Equation as

L.H.S. = $(\sqrt{a} +\sqrt{b})(\sqrt{a}-\sqrt{b})$

R.H.S. = $a-b$

Since R.H.S. is itself in simplified form so we need to work on L.H.S. which can be done as follows

$L.H.S.=(\sqrt{a} +\sqrt{b})(\sqrt{a}-\sqrt{b})\\ =\sqrt{a}.\sqrt{a}-\sqrt{a}.\sqrt{b}+\sqrt{b}.\sqrt{a}-\sqrt{b}.\sqrt{b}\\=a+\sqrt{ab}+\sqrt{ab} -b$

⇒$L.H.S.=a-b$

Hence L.H.S. = R.H.S.

So we have proved the expression is given as $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$