Question

find the value of x.​

Answer 1

Answer :-

Value of x is (C) 3.

Explanation :-

$\mathsf{ \{ x^2 + (x + 1)^2 \}^{ - \dfrac{1}{2} }= (x + 2)^{ - 1} }$

$\mathsf{ \implies \{ x^2 + (x^2 + 2(1)(x) + 1^2 \}^{ - \dfrac{1}{2} }= (x + 2)^{ - 1} }$

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

$\mathsf{ \implies \{ x^2 + x^2 + 2x + 1 \}^{ - \dfrac{1}{2} }= (x + 2)^{ - 1} }$

$\mathsf{ \implies \{ 2x^2+ 2x + 1 \}^{ - \dfrac{1}{2} }= (x + 2)^{ - 1} }$

Squaring on both sides

$\mathsf{ \implies \{ 2x^2+ 2x + 1 \}^{ - 1}= (x + 2)^{ - 2}}$

$\mathsf{ \implies \dfrac{1}{\{ 2x^2+ 2x + 1 \}^{1}}= \dfrac{1}{(x + 2)^{2}}}$

$\boxed{ \bf \because a^{ - m} = \dfrac{1}{ {a}^{m} }}$

$\mathsf{ \implies \dfrac{1}{2x^2+ 2x + 1} = \dfrac{1}{(x + 2)^{2}}}$

By cross multiplication

$\mathsf{ \implies {2x^2+ 2x + 1} = {(x + 2)^{2}}}$

$\mathsf{ \implies {2x^2+ 2x + 1} = {x}^{2} + 2(x)(2) + {2}^{2} }$

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

$\mathsf{ \implies {2x^2+ 2x + 1} = {x}^{2} + 4x+ 4 }$

$\mathsf{ \implies {2x^2+ 2x + 1} - {x}^{2} - 4x - 4 = 0}$

$\mathsf{ \implies x^2 - 2x - 3 = 0}$

$\mathsf{ \implies x^2 - 3x + x - 3 = 0}$

$\mathsf{ \implies x(x - 3) + 1( x - 3) = 0}$

$\mathsf{ \implies (x - 3)(x + 1) = 0}$

$\mathsf{ \implies x - 3 = 0 \ or \ x + 1= 0}$

$\mathsf{ \implies x = 3 \ or \ x = - 1}$

$\mathsf{ \implies x = 3 \qquad \{neglecting \ x = - 1 \}}$

∴ the value of x is (C) 3.

Answer 2

Given

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

To find

The value of x

Explanation

consider

$\sf {x^2+(x+1)^2}^{\dfrac{-1}{2}}=(x+2)^{-1}$

squaring on both sides

$\sf \: (({x}^{2} + {(x + 1)}^{2} )^{ \dfrac{ - 1}{2} } ) ^{2} = ( {(x + 2)}^{ - 1} ) ^{2}$

we know that

$\boxed {\bf \: {({a}^{m}) } ^{n} = {a}^{mn} }$

$\sf \: ( {x}^{2} + {(x + 1)}^{2} ) ^{ - 1} = {(x + 2)}^{ - 2}$

$\boxed{ \bf \: {a}^{ - m} = \dfrac{1}{ {a}^{m} } }$

$\sf \: = > \dfrac{1}{ {x}^{2} + {(x + 1)}^{2} } = \dfrac{1}{ {(x + 2)}^{2} }$

$\boxed{ \bf \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

$\sf \: \dfrac{1}{{x}^{2} + {x}^{2} + 1 + 2x} = \dfrac{1}{ {x}^{2} + 4 + 4x } \\ \\ = > \sf \: \dfrac{1}{2 {x}^{2} + 2x + 1} = \dfrac{1}{ {x}^{2} + 4x + 4 }$

Cross multiplying

$= > \sf {x}^{2} + 4x + 4 = 2 {x}^{2} + 2x + 1 \\ \\ \sf \: = > {x}^{2} - 2x - 3 = 0$

splitting the middle term

$\sf \: = > {x}^{2} - 3x + x - 3 = 0 \\ \\ \sf = > x(x - 3) + 1(x - 3) = 0$

$\sf \: = > (x - 3)(x + 1) = 0$

$\sf \: x = - 1 \: and \: x = 3$

ignoring x=-1

$\boxed{ \huge \sf \: x = 3}$

More formulas

➡In an quadratic equation

if the roots are not real

then,

$\bullet \sf x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

➡$\sf Product\: of\: roots =\dfrac{constant \:term}{coefficient\:of\: x^2}$

➡$\sf sum \:of \:roots =\dfrac{-coefficient \:of \:x}{coefficient\: of \:x^2}$