Question

If x = 1/3+root 7 then prove that 2x^2 – 6x + 1 = 0

Answer 1

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\:\boxed{ \rm{ \: x = \frac{1}{3 + \sqrt{7} } \: \: }}}$

$\large\underline{\sf{To\:prove - }}$

$\red{\rm :\longmapsto\:\boxed{ \rm{ \: \: {2x}^{2} - 6x + 1 = 0 \: \: \: }}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = \dfrac{1}{3 + \sqrt{7} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{1}{3 + \sqrt{7} } \times \red{\dfrac{3 - \sqrt{7} }{3 - \sqrt{7} } }$

We know,

$\purple{\boxed{ \rm{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:x = \dfrac{3 - \sqrt{7} }{ {(3)}^{2} - {( \sqrt{7}) }^{2} }$

$\rm :\longmapsto\:x = \dfrac{3 - \sqrt{7} }{9 - 7}$

$\rm :\longmapsto\:x = \dfrac{3 - \sqrt{7} }{2}$

$\rm :\longmapsto\:2x = 3 - \sqrt{7}$

$\rm :\longmapsto\:2x - 3= - \sqrt{7}$

On squaring both sides, we get

$\rm :\longmapsto\: {(2x - 3)}^{2} = {( - \sqrt{7})}^{2}$

$\rm :\longmapsto\: {(2x)}^{2} + {(3)}^{2} - 2(2x)(3) = 7$

$\rm :\longmapsto\: {4x}^{2} + 9 - 12x = 7$

$\rm :\longmapsto\: {4x}^{2} + 9 - 12x - 7 = 0$

$\rm :\longmapsto\: {4x}^{2} - 12x + 2 = 0$

$\rm :\longmapsto\:2( {2x}^{2} - 6x + 1) = 0$

$\bf\implies \: {2x}^{2} - 6x + 1 = 0$

Hence, Proved

More Identities to know :-

$\red{\boxed{ \rm{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

$\blue{\boxed{ \rm{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

$\green{\boxed{ \rm{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: \: }}}$

$\orange{\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) \: \: }}}$

$\purple{\boxed{ \rm{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: \: }}}$

$\pink{\boxed{ \rm{ \: {(x + y)}^{2} + {(x - y)}^{2} =2[ \: {x}^{2} + {y}^{2} \: ] \: \: }}}$

$\red{\boxed{ \rm{ \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}}$

Answer 2

Answer:

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\:\boxed{ \rm{ \: x = \frac{1}{3 + \sqrt{7} } \: \: }}}$

$\large\underline{\sf{To\:prove - }}$

$\red{\rm :\longmapsto\:\boxed{ \rm{ \: \: {2x}^{2} - 6x + 1 = 0 \: \: \: }}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = \dfrac{1}{3 + \sqrt{7} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{1}{3 + \sqrt{7} } \times \red{\dfrac{3 - \sqrt{7} }{3 - \sqrt{7} } }$

We know,

$\purple{\boxed{ \rm{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:x = \dfrac{3 - \sqrt{7} }{ {(3)}^{2} - {( \sqrt{7}) }^{2} }$

$\rm :\longmapsto\:x = \dfrac{3 - \sqrt{7} }{9 - 7}$

$\rm :\longmapsto\:x = \dfrac{3 - \sqrt{7} }{2}$

$\rm :\longmapsto\:2x = 3 - \sqrt{7}$

$\rm :\longmapsto\:2x - 3= - \sqrt{7}$

On squaring both sides, we get

$\rm :\longmapsto\: {(2x - 3)}^{2} = {( - \sqrt{7})}^{2}$

$\rm :\longmapsto\: {(2x)}^{2} + {(3)}^{2} - 2(2x)(3) = 7$

$\rm :\longmapsto\: {4x}^{2} + 9 - 12x = 7$

$\rm :\longmapsto\: {4x}^{2} + 9 - 12x - 7 = 0$

$\rm :\longmapsto\: {4x}^{2} - 12x + 2 = 0$

$\rm :\longmapsto\:2( {2x}^{2} - 6x + 1) = 0$

$\bf\implies \: {2x}^{2} - 6x + 1 = 0$

Hence, Proved

More Identities to know :-

$\red{\boxed{ \rm{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

$\blue{\boxed{ \rm{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

$\green{\boxed{ \rm{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: \: }}}$

$\orange{\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) \: \: }}}$

$\purple{\boxed{ \rm{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: \: }}}$

$\pink{\boxed{ \rm{ \: {(x + y)}^{2} + {(x - y)}^{2} =2[ \: {x}^{2} + {y}^{2} \: ] \: \: }}}$

$\red{\boxed{ \rm{ \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}}$