Question

Solve each equation. Check your solutions.

$\mathbf{4}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{64}$

Short answer

The solution of the given equation is:

$\mathit{c}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{1}$

Step-by-step solution

Step 1. State the concept for quadratic equation.

A quadratic equation is a second order equation written as$a{x}^2+bx+c=0$ where a, b, and c are coefficients of real numbers and $a\ne 0$.

A quadratic equation generally has two roots, which can be equal to each other. Some quadratic expressions can be factored, which means that the equation can be written as:

$a\left(x-x_1\right)\left(x-x_2\right)=0$

where$x_1$ and$x_2$ are the two roots of the equation.

Step 2. State the methods to factorize.

Factoring an algebraic expression means writing the expression as a product of factors.

To verify whether the factors are correct or not, multiply them and check if the result is the original algebraic expression.

Algebraic expressions can be factorized using the common factor method, regrouping like terms together, and also by using algebraic identities.

Step 3. Solve the equation.

Factorize the given polynomial to solve the equation.

$\begin{array}{c}4y^2=64\\ 4y^2-64=0\\ {\left(2y\right)}^2-8^2=0\\ \left(2y-8\right)\left(2y+8\right)=0\\ 2y-8=0\text{or}2y+8=0\\ y=4\text{or}y=-4\\ y=4,-4\end{array}$

Therefore, the solution to the given equation is $y=4,-4$.

Step 4. Check the solution.

Put the values in the equation to check if the solution is correct.

If the solution is correct, LHS = RHS.

$\begin{array}{c}\text{for}y=-4\\ 4y^2=64\\ 4{\left(-4\right)}^2=64\\ 4\left(16\right)=64\\ 64=64\\ LHS=RHS\end{array}$

$\begin{array}{c}\text{for}y=4\\ 4y^2=64\\ 4{\left(4\right)}^2=64\\ 4\left(16\right)=64\\ 64=64\\ LHS=RHS\end{array}$

Therefore, the solution is correct.