Question

How do you solve the equation $4{{\left( x+1 \right)}^{2}}=64$ .

Answer 1

Hint: Now consider the given equation. First we will divide the whole equation by 4. Now we will further take square root on both sides and simplify the equation. Now on simplification we will have two linear equations in one variable. We will solve linear equations to get the solution to the equation.

Complete step by step solution:
Now first consider the given equation $4{{\left( x+1 \right)}^{2}}=64$ .
Now to solve the equation we will first separate the square term. Hence we will divide the equation by 4. On dividing we get,
$\begin{align}
  & \Rightarrow {{\left( x+1 \right)}^{2}}=\dfrac{64}{4} \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}=16 \\
\end{align}$
Now since we have square on one side we will simplify using square root. Hence taking square root on both sides we get,
$\begin{align}
  & \Rightarrow \left( x+1 \right)=\pm \sqrt{16} \\
 & \Rightarrow \left( x+1 \right)=\pm 4 \\
\end{align}$
Hence we have x + 1 = 4 or x + 1 = - 4.
Now we have two linear equations in one variable. We will solve these equations to find the value of x.
First consider the equation x + 1 = 4.
Now taking 1 on RHS we get, x = 4 – 1.
Hence x = 3.
Now consider the equation x + 1 = - 4.
Hence taking 1 on RHS we get x = - 4 - 1.
Hence x = - 5.
Hence the solution of the given equation is x = 3 or x = -5.

Note:
Now note that since we have square the equation formed will be quadratic. We can also open the square by the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Now we write the expression in the form of $a{{x}^{2}}+bx+c=0$ and we know that the roots of this quadratic equation is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . hence we have the solution of the given equation.