Question

2. Evaluate :
(i) (1+√x)^2+(1-√x)^2​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$2. Evaluate :

(i) (1+√x)^2+(1-√x)^2

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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Formula used here :

$\bold{\boxed{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\bold{\boxed{{(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

Now,Come to the question:

$= > {(1 + \sqrt{x}) }^{2} = {1}^{2} + { (\sqrt{x} )}^{2} + 2 \sqrt{x}$

$= > {(1 - \sqrt{x}) }^{2} = {1}^{2} + {( \sqrt{x} )}^{2} - 2 \sqrt{x}$

Sum:-

$= > {(1 + \sqrt{x} )}^{2} + {(1 - \sqrt{x} )}^{2} = 1 + {( \sqrt{x}) }^{2} + 2 \sqrt{x} + 1 + {( \sqrt{x}) }^{2} - 2 \sqrt{x}$

$= 1 + x + {\cancel{2 \sqrt{x} }} + 1 + x - {\cancel{2 \sqrt{x}}}$

$\bold{\pink{= 2 + 2x = 2(1 + x)}}$

$\huge\mathcal{Other\: Formulas:}$

1. $\bold{\boxed{{(x  ±y)}^{3} = {x}^{3} +   {y}^{3} ±(x ±y)}}$
2. $\bold{\boxed{ {x}^{3}   + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)}}$
3. $\bold{\boxed{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)}}$
4. $\bold{\boxed{ {x}^{2} - {y}^{2} = (x + y)(x - y)}}$

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