Question

How do you solve \[\dfrac{2}{3}{\left( {x + 8} \right)^2} - 66 = 0\]?

Answer 1

Hint: In order to determine solution of the above equation , multiply both sides of the equation with the number $\dfrac{3}{2}$and simply it . Now transpose the constant term on the right-hand side of the equation and take the square root on both sides and solve it for $x$to obtain the required solution.

Complete step-by-step solution:
Given a quadratic equation, \[\dfrac{2}{3}{\left( {x + 8} \right)^2} - 66 = 0\] let it be $f(x)$
$f(x) = \dfrac{2}{3}{\left( {x + 8} \right)^2} - 66 = 0$
First we will be simplifying our equation by multiplying both sides with $\dfrac{3}{2}$
\[\Rightarrow\dfrac{3}{2}\left( {\dfrac{2}{3}{{\left( {x + 8} \right)}^2} - 66} \right) = 0 \times \dfrac{3}{2}\]
Using distributive law for multiplication $A\left( {B + C} \right) = AB + AC$
\[\Rightarrow \dfrac{3}{2}\left( {\dfrac{2}{3}{{\left( {x + 8} \right)}^2}} \right) - \dfrac{3}{2}\left( {66} \right) = 0\]
Simplifying further we get
\[\Rightarrow {\left( {x + 8} \right)^2} - 99 = 0\]
Now transposing the constant terms on the right-hand side of the equation
\[\Rightarrow {\left( {x + 8} \right)^2} = 99\]
Taking square root on both sides
\[
\Rightarrow x + 8 = \pm \sqrt {99} \\
\Rightarrow x = - 8 \pm \sqrt {99} \\
 \]
Therefore , solution to the equation \[\dfrac{2}{3}{\left( {x + 8} \right)^2} - 66 = 0\]is \[x = - 8 \pm \sqrt {99} \].

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$. If $a = 0$ then the equation will become a linear equation and will no longer be quadratic.
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
-If D is equal to zero, then both of the roots will be the same and real.
-If D is a positive number then, both of the roots are real solutions.
-If D is a negative number, then the root are the pair of complex solutions

Note:
1. Must be careful while calculating the answer as calculation error may come.
2.Always place $ \pm $on the other side of the equation .
3. Whenever we transpose any term from one side to another side the sign of the term gets reversed.