Question

What is the solution of the equation ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} + {\left( {3 + 2\sqrt 2 } \right)^{8 - {x^2}}} = 6$?
A. $3 \pm 2\sqrt 2 $
B. $ \pm 1$
C. $ \pm 3\sqrt 3 , \pm 2\sqrt 2 $
D. $ \pm 7, \pm \sqrt 3 $

Answer 1

Hint: First we will assign variables to the expression ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}}$. Then we will solve the equation. After that, we substitute the value of the variable and calculate the value of $x$.

Formula Used:
The solution of the quadratic equation $a{x^2} + bx + c = 0$ is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step by step solution:
Given equation is ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} + {\left( {3 + 2\sqrt 2 } \right)^{8 - {x^2}}} = 6$.
Rewrite the given equation:
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} + \dfrac{1}{{{{\left( {3 + 2\sqrt 2 } \right)}^{{x^2} - 8}}}} = 6$ …..(1)
Assume that ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = t$
Substitute ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = t$ in the equation (1)
$ \Rightarrow t + \dfrac{1}{t} = 6$
Simplify the left side expression of the equation
$ \Rightarrow \dfrac{{{t^2} + 1}}{t} = 6$
Multiply $t$ both sides
$ \Rightarrow {t^2} + 1 = 6t$
Subtract $6t$ from both sides
$ \Rightarrow {t^2} - 6t + 1 = 0$
Now compare the above equation with the standard form of the quadratic equation $a{x^2} + bx + c = 0$.
$a = 1, b = - 6, c = 1$
Now apply the quadratic formula to solve ${t^2} - 6t + 1 = 0$
$ \Rightarrow t = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4 \cdot 1 \cdot 1} }}{{2 \cdot 1}}$
$ \Rightarrow t = \dfrac{{6 \pm \sqrt {36 - 4} }}{2}$
$ \Rightarrow t = \dfrac{{6 \pm \sqrt {32} }}{2}$
Rewrite $\sqrt {32} $ as $4\sqrt 2 $ since $32 = 16 \times 2$.
$ \Rightarrow t = \dfrac{{6 \pm 4\sqrt 2 }}{2}$
Take common 2 from numerator
$ \Rightarrow t = \dfrac{{2\left( {3 \pm 2\sqrt 2 } \right)}}{2}$
Cancel out 2 from the denominator and numerator.
$ \Rightarrow t = 3 \pm 2\sqrt 2 $
Now put the value of $t$.
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = 3 \pm 2\sqrt 2 $
Solve the equation ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \left( {3 + 2\sqrt 2 } \right)$
${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = {\left( {3 + 2\sqrt 2 } \right)^1}$
Compare the power of expression
$ \Rightarrow {x^2} - 8 = 1$
Add 8 on both sides of equation
$ \Rightarrow {x^2} = 9$
$ \Rightarrow x = \pm 3$
Solve the equation ${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \left( {3 - 2\sqrt 2 } \right)$
${\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \left( {3 - 2\sqrt 2 } \right)$
Multiply $\dfrac{{\left( {3 + 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)}}$ with the right-side expression.
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \left( {3 - 2\sqrt 2 } \right) \times \dfrac{{\left( {3 + 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)}}$
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \dfrac{{\left( {3 - 2\sqrt 2 } \right)\left( {3 + 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)}}$
Apply the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
 $ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \dfrac{{{{\left( 3 \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}}{{\left( {3 + 2\sqrt 2 } \right)}}$
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \dfrac{{9 - 8}}{{\left( {3 + 2\sqrt 2 } \right)}}$
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = \dfrac{1}{{\left( {3 + 2\sqrt 2 } \right)}}$
$ \Rightarrow {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} = {\left( {3 + 2\sqrt 2 } \right)^{ - 1}}$
Compare the power of the expression
$ \Rightarrow {x^2} - 8 = - 1$
$ \Rightarrow {x^2} = - 1 + 8$
$ \Rightarrow {x^2} = 7$
$ \Rightarrow x = \pm \sqrt 7 $
Thus the roots of $x$ is $ \pm 3$ and $ \pm \sqrt 7 $.

Option ‘D’ is correct

Note: We can apply quadratic formula to solve ${x^2} - 8 = - 1$ an ${x^2} - 8 = 1$. Therefore, the solution of ${x^2} - 7 = 0$ by using quadratic formula is $x = \dfrac{{ - 0 \pm \sqrt {{{\left( 0 \right)}^2} - 4 \cdot 1 \cdot \left( { - 7} \right)} }}{{2 \cdot 1}} \Rightarrow x = \pm \sqrt 7 $. . Therefore, the solution of ${x^2} - 9 = 0$ by using quadratic formula is $x = \dfrac{{ - 0 \pm \sqrt {{{\left( 0 \right)}^2} - 4 \cdot 1 \cdot \left( { - 9} \right)} }}{{2 \cdot 1}} \Rightarrow x = \pm 3$.