Question

The solution set of equation ${(5 + 2\sqrt 6 )^{{x^2} - 3}} + {(5 - 2\sqrt 6 )^{{x^2} - 3}} = 10$ is
(A) \left\\{ { \pm {\text{ }}2, \pm {\text{ }}\sqrt 2 } \right\\}
(B) \left\\{ { \pm {\text{ 3}}, \pm {\text{ }}\sqrt 3 } \right\\}
(C) \left\\{ { \pm {\text{ 5}}, \pm {\text{ }}\sqrt 5 } \right\\}
(D) \left\\{ { \pm {\text{ 6}}, \pm {\text{ }}\sqrt 6 } \right\\}

Answer 1

To solve this question we use the concepts of quadratic equations. And also, we use some basic level math to solve this question. A quadratic equation is a second degree equation in one variable, which has two roots. It will be of the form, $a{x^2} + bx + c = 0$ .
Formula used:
${\text{roots of the equation }}a{x^2} + bx + c = 0{\text{ is }}x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . Here the phrase inside the root is called a determinant, which decides the nature of the roots. If the determinant is less than 0, then roots are imaginary. If the determinant is more than zero, roots are real and distinct. If the determinant is equal to zero, roots are real but equal.

Complete step-by-step solution:
We know that $(a + b)(a - b) = {a^2} - {b^2}$
So (5 + 2\sqrt 6 )(5 - 2\sqrt 6 ) = {5^2} - {(2\sqrt 6 )^2}$$$$ = 25 - 24 = 1
So, $5 - 2\sqrt 6 = \dfrac{1}{{5 + 2\sqrt 6 }}$ -----(1)
So we get the equation as ${(5 + 2\sqrt 6 )^{{x^2} - 3}} + \dfrac{1}{{{{(5 + 2\sqrt 6 )}^{{x^2} - 3}}}} = 10$
Let's simplify the equations, as it is a complicated process solving these equations. So we will replace the complicated phrase with a simple variable. Now, we can change the equation as $n + \dfrac{1}{n} = 10$ , where, $n = {(5 + 2\sqrt 6 )^{{x^2} - 3}}$
Now let's solve the equation $n + \dfrac{1}{n} = 10$
On multiplying this whole equation by $n$ , we get, ${n^2} - 10n + 1 = 0$
So by using quadratic formula, we get $n = \dfrac{{10 \pm \sqrt {100 - 4} }}{2}$
So $n = 5 \pm 2\sqrt 6$
But we know the fact that the variable $n = {(5 + 2\sqrt 6 )^{{x^2} - 3}}$
So ${(5 + 2\sqrt 6 )^{{x^2} - 3}} = 5 \pm 2\sqrt 6$
$\Rightarrow {(5 + 2\sqrt 6 )^{{x^2} - 3}} = 5 + 2\sqrt 6 {\text{ or }}{(5 + 2\sqrt 6 )^{{x^2} - 3}} = 5 - 2\sqrt 6$
$\Rightarrow {(5 + 2\sqrt 6 )^{{x^2} - 3}} = {(5 + 2\sqrt 6 {\text{)}}^1}{\text{ or }}{(5 + 2\sqrt 6 )^{{x^2} - 3}} = \dfrac{1}{{5 + 2\sqrt 6 }} = {(5 + 2\sqrt 6 )^{ - 1}}$ ------from eq.(1)
So we conclude that ${x^2} - 3 = 1{\text{ or }}{x^2} - 3 = - 1$
$\Rightarrow {x^2} = 4{\text{ or }}{x^2} = 2$
$\Rightarrow x = \pm 2{\text{ or }}x = \pm \sqrt 2$
So option (A) is correct.

Note: In all the equations, both positive and negative values are considered so that we get all the possible solutions. The quadratic equation can also be solved using completing the square method also. Make sure that you consider all the values in every equation. And be careful while solving the surds or the equations in which the roots are included.