Question

If $\mathrm{x=\dfrac{2}{3+\sqrt7}}$, then $\mathrm{(x-3)^2}$ is = _____

[a] 7
[b] 6
[c] 1
[d] 3

Answer 1

Required answer:-

7 (option -a)

Given :-

$x =\dfrac{2}{3+\sqrt{7} }$

To find :-

$(x-3)^2$

Solution :-

First we rationalize the denominator of the x

$x =\dfrac{2}{3+\sqrt{7} }$

In order to rationalize the denominator multiply and divide with its conjugate

The conjugate of $3+\sqrt{7}$ is $3 -\sqrt{7}$ So, multiply and divide with this

$x =\dfrac{2}{3+\sqrt{7} } \times\dfrac{3-\sqrt{7} }{3-\sqrt{7} }$

$x =\dfrac{2(3-\sqrt{7}) }{(3+\sqrt{7} )(3-\sqrt{7}) }$

Denominator can be simplified by using algebraic identity

(a+b)(a-b)= a^2-b^2

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

$x =\dfrac{2(3-\sqrt{7} ) }{9-7}$

$x =\dfrac{2(3-\sqrt{7} ) }{2}$

${\boxed{x = 3-\sqrt{7} }}$

Substitute value of x in (x-3)^2 in order to get the required value

$(x-3)^2$

$(3-\sqrt{7} -3)^2$

$(\not3-\sqrt{7} \not-3)^2$

$(-\sqrt{7} )^2$

$=7$

So,

${\boxed{If x=\dfrac{2}{3+\sqrt7} then, (x-3)^2 = 7}}$

So, the correct option is {a}

Know more algebraic identities :-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc