Question

How do you solve ${(2x + 4)^2} = 64$ ?

Answer 1

Hint: Firstly, write the squared term twice and then multiply them. Rearrange the terms to get a quadratic equation and then use the roots for the quadratic equation formula to get the roots. The roots will be the solution to the equation.

Formula used: The roots of a quadratic equation( $a{x^2} + bx + c = 0$ ) can be found by the formula,
$x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$

Complete step-by-step solution:
The given expression is, ${(2x + 4)^2} = 64$
Expand the squared term.
$\Rightarrow (2x + 4)(2x + 4) = 64$
Now multiply both the terms and write the equation again.
$\Rightarrow [2x(2x + 4) + 4(2x + 4)] = 64$
$\Rightarrow [(4{x^2} + 8x) + (8x + 16)] = 64$
Now add all the same degree terms.
$\Rightarrow 4{x^2} + 16x + 16 = 64$
Divide the entire equation $4$ for simplification purposes.
$\Rightarrow \dfrac{1}{4}(4{x^2} + 16x + 16) = \dfrac{{64}}{4}$
Simplify further,
$\Rightarrow {x^2} + 4x + 4 = 16$
Subtract $\;16$ from both the sides of the equation.
$\Rightarrow {x^2} + 4x + 4 - 16 = 16 - 16$
$\Rightarrow {x^2} + 4x - 12 = 0$
Now use the formula for finding the roots of the above quadratic expression.
The formula is $x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$
Here, $a = 1;b = 4;c = - 12$
On substituting these values in the formula, we get,
$\Rightarrow x = \left[ {\dfrac{{ - 4 \pm \sqrt {{4^2} - 4 \times 1 \times ( - 12)} }}{{2 \times 1}}} \right]$
First, solve the operations in the square root.
$\Rightarrow x = \left[ {\dfrac{{ - 4 \pm \sqrt {16 + 48} }}{{2 \times 1}}} \right]$
$\Rightarrow x = \left[ {\dfrac{{ - 4 \pm \sqrt {64} }}{{2 \times 1}}} \right]$
Write the values of the square root.
$\Rightarrow x = \left[ {\dfrac{{ - 4 \pm 8}}{2}} \right]$
Simplify further.
$\Rightarrow x = \left[ { - 2 \pm 4} \right]$
Open the $\pm$ sign to get two values of $x$
$\Rightarrow x = - 2 + 4;x = - 2 - 4$
Evaluate them
$\Rightarrow x = 2;x = - 6$

Hence the roots of the expression, ${(2x + 4)^2} = 64$ are $x = 2, - 6$.

Additional Information: A polynomial is a mathematical expression that contains one or more variables in sum or subtraction format with different powers. Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial.

Note: Never forget to take two conditions whenever there is a $\pm$ symbol. The expression is written twice once with $+\;$ and another time with $\;-$ . You can always cross-check your answer by placing the value of $x$ back in the equation. If you get LHS = RHS then your answer is correct.