Question

the values of x for the equations square root of (2x²+5x-2)- square root of (2x²+5x+9) =1​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\rm : \implies \: \sqrt{ {2x}^{2} + 5x - 2} - \sqrt{ {2x}^{2} + 5x + 9} = 1$

$\boxed{ \pink{ \rm : \implies \:Let \: {2x}^{2} + 5x = y }} - - - (i)$

So, given equation can be rewritten as

$\rm : \implies \: \sqrt{y - 2} - \sqrt{y + 9} = 1$

$\rm : \implies \: \sqrt{y - 2} = 1 + \sqrt{y + 9}$

$\green{ \rm \: squaring \: both \: sides}$

$\rm : \implies \: \cancel{y} - 2 = 1 + \cancel{y} + 9 + 2 \sqrt{y + 9}$

$\rm : \implies \: - \: 2 \sqrt{y + 9} = 12$

$\rm : \implies \: \sqrt{y + 9} = - \: 6$

$\green{ \rm \: squaring \: both \: sides}$

$\rm : \implies \:y + 9 = 36$

$\rm : \implies \:y = 27$

On substituting the value of y from equation (i), we get

$\red{ \rm : \implies \: {2x}^{2} + 5x = 27}$

$\rm : \implies \: {2x}^{2} + 5x - 27 = 0$

$\rm : \implies \: on \: comparing \: with \: {ax}^{2} + bx + c = 0 \: we \: get$

a = 2

b = 5

c = - 27

Therefore, Discriminant, D is given by

$\rm : \implies \:D \: = \: {b}^{2} - 4ac$

$\rm : \implies \:D = {5}^{2} - 2 \times 2 \times ( - 27)$

$\rm : \implies \:D \: = \: 25 + 108$

$\rm : \implies \:D = 133$

Hence, solution is given by

$\pink{ \: \rm : \implies \:x \: = \dfrac{ - b \: \pm \: \sqrt{ D} }{2a} }$

$\rm : \implies \:x \: = \: \dfrac{ - 5 \: \pm \: \sqrt{133} }{4}$

$\rm : \implies \:x = \dfrac{ - 5 \: \pm \: 11.5}{4}$

$\rm : \implies \:x \: = \: \dfrac{ - 5 + 11.5}{4} \: \: or \: \: \dfrac{ - 5 - 11.5}{4}$

$\rm : \implies \:x \: = \: \dfrac{5.5}{4} \: \: or \: \: \dfrac{ - 16.5}{4}$

$\rm : \implies \:x \: = \: 1.625 \: \: or \: \: - \: 4.125$

Verification :-

When y = 27,

Consider, LHS

$\rm : \implies \: \sqrt{y - 2} - \sqrt{y + 9}$

$\rm : \implies \: \sqrt{27 - 2} - \sqrt{27 + 9}$

$\rm : \implies \: \sqrt{25} - \sqrt{36}$

$\rm : \implies \:5 - 6$

$\rm : \implies \: - 1$

$\rm : \implies \: \ne \: RHS$

Hence, given equation has no solution.