How do you solve ${\left( {x + 4} \right)^2} = 81$?

Hint: As the given equation is quadratic in one variable, we will use the quadratic formula to find the roots of the given equation. First, expand the left side by the formula, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. After that move the constant part on the left side. If the given quadratic equation is of the form $a{x^2} + bx + c = 0$, the quadratic formula is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step by step answer:
We have been given an equation ${\left( {x + 4} \right)^2} = 81$.
We have to find the roots of the given equation.
Now, expand the term by using the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
$ \Rightarrow {x^2} + 8x + 16 = 81$
Move the constant part on the left side and simplify,
$ \Rightarrow {x^2} + 8x - 65 = 0$
Now, we will compare the given equation with the standard quadratic equation which is given by $a{x^2} + bx + c = 0$.
On comparing we get the values $a = 1,b = 8,c = - 65$.
Now, we know that the quadratic formula is given as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values in the above formula we get
$ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {{8^2} - 4 \times 1 \times - 65} }}{{2 \times 1}}$
Now, on solving the obtained equation we get
$ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {64 + 260} }}{2}$
Add the terms in the square root,
$ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {324} }}{2}$
Now, we know that the value of the square root $\sqrt {324} = 18$.
$ \Rightarrow x = \dfrac{{ - 8 \pm 18}}{2}$
Take 2 commons from the numerator,
$ \Rightarrow x = \dfrac{{2\left( { - 4 \pm 9} \right)}}{2}$
Cancel out the common factors,
$ \Rightarrow x = - 4 \pm 9$
Now, we know that a quadratic equation has two roots. We can write the obtained equation as
$ \Rightarrow x = - 4 - 9$ and $x = - 4 + 9$
Simplify the terms,
$ \Rightarrow x = - 13$ and $x = 5$

Hence, the roots of the equation ${\left( {x + 4} \right)^2} = 81$ is $ - 13$ and $5$.

Note: Avoid calculation mistakes because single calculation mistakes lead to an incorrect answer. To solve a quadratic equation students can use the factorization method, completing the square method or quadratic formula method. When the time is less and we are sure about the quadratic formula, then it is best to use this method. We can cross verify the factors by opening parenthesis and solving.