Answer

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**Hint:**In this question take ${2^x} = y$ which will help you to obtain a new quadratic equation in terms of y to find the roots of the new quadratic equation using the factoring method, using the value of roots you can obtain the value of x and easier to identify the correct option.

**Complete step-by-step solution:**According to the given information we have equation ${2^{2x + 3}} - 57 = 65\left( {{2^x} - 1} \right)$

Since we know that ${a^{n + m}} = {a^n} \times {a^m}$

Therefore, the above equation can be re-written as; ${\left( {{2^x}} \right)^2} \times {2^3} = 65\left( {{2^x} - 1} \right) + 57$................ (equation 1)

let ${2^x} = y$ .......... (equation 2)

substituting the y in the equation 1 we get

${y^2} \times {2^3} = 65\left( {y - 1} \right) + 57$

$ \Rightarrow $$8{y^2} = 65y - 65 + 57$

$ \Rightarrow $$8{y^2} - 65y + 8 = 0$

Now using the factoring method to find the roots of the above quadratic equation we get

$8{y^2} - \left( {64 + 1} \right)y + 8 = 0$

$ \Rightarrow $$8{y^2} - \left( {64 + 1} \right)y + 8 = 0$

$ \Rightarrow $$8{y^2} - 64y - 1y + 8 = 0$

$ \Rightarrow $$8y\left( {y - 8} \right) - 1\left( {y - 8} \right) = 0$

$ \Rightarrow $$\left( {8y - 1} \right)\left( {y - 8} \right) = 0$

So, $y = 8$ or $y = \dfrac{1}{8}$

Now substituting the value of y in the equation 2

For y = 8

$\Rightarrow {2^x} = 8$

$\Rightarrow {2^x} = {\left( 2 \right)^3}$

Therefore, x = 3

Now for $y = \dfrac{1}{8}$

$\Rightarrow {2^x} = \dfrac{1}{8}$

$\Rightarrow {2^x} = {\left( 2 \right)^{ - 3}}$

Therefore, $x = – 3 $

So, the roots of the given is $ \pm 3$

**Hence, option B is the correct option.**

**Note:**In the above solution we used a term “quadratic equation” which can be defined as the polynomial equation of degree two which implies that this equation has only one term with highest power two. A quadratic equation general form is represented as $a{x^2} + bx + c = 0$ here x is variable which is unknown and a, b, c are the numerical coefficients. The unknown values of x can be find using the quadratic formula which is given as; \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

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