Respuesta :
Answer
The solutions to the quadratic equations are
[tex]\begin{gathered} a^2-4a-45 \\ \text{Solution: }a=-5\text{ or }9 \\ \\ 5y^2+4y=0 \\ \text{Solution: }y=0\text{ or }-\frac{4}{5} \end{gathered}[/tex]SOLUTION
Problem Statement
The question gives us 2 quadratic equations and we are required to solve them by factoring, first writing them in their standard forms.
The quadratic equations given are:
[tex]\begin{gathered} a^2-4a-45=0 \\ 5y^2+4y=0 \end{gathered}[/tex]Method
To solve the questions, we need to follow these steps:
(We will represent the independent variable as x for this explanation. We know they are "a" and "y" in the questions given)
The steps outlined below are known as the method of Completing the Square.
Step 1: Find the square of the half of the coefficient of x.
Step 2: Add and subtract the result from step 1.
Step 3: Re-write the Equation. This will be the standard form of the equation
Step 4. Solve for x
We will apply these steps to solve both questions.
Implementation
Question 1:
[tex]\begin{gathered} a^2-4a-45=0 \\ \text{Step 1: Find the square of the half of the coefficient of }a \\ (-\frac{4}{2})^2=(-2)^2=4 \\ \\ \text{Step 2: Add and subtract 4 to the equation} \\ a^2-4a-45+4-4=0 \\ \\ \text{Step 3: Rewrite the Equation} \\ a^2-4a+4-45-4=0 \\ (a^2-4a+4)-49=0 \\ (a^2-4a+4)=(a-2)^2 \\ \therefore(a-2)^2-49=0 \\ \text{ In standard form, we have:} \\ (a-2)^2=49 \\ \\ \text{Step 4: Solve for }a \\ (a-2)^2=49 \\ \text{ Find the square root of both sides} \\ \sqrt[]{(a-2)^2}=\pm\sqrt[]{49} \\ a-2=\pm7 \\ \text{Add 2 to both sides} \\ \therefore a=2\pm7 \\ \\ \therefore a=-5\text{ or }9 \end{gathered}[/tex]Question 2:
[tex]\begin{gathered} 5y^2+4y=0 \\ \text{ Before we begin solving, we should factorize out 5} \\ 5(y^2+\frac{4}{5}y)=0 \\ \\ \text{Step 1: Find the square of the coefficient of the half of y} \\ (\frac{4}{5}\times\frac{1}{2})^2=(\frac{2}{5})^2=\frac{4}{25} \\ \\ \text{Step 2: Add and subtract }\frac{4}{25}\text{ to the equation} \\ \\ 5(y^2+\frac{4}{5}y+\frac{4}{25}-\frac{4}{25})=0 \\ \\ \\ \text{Step 3: Rewrite the Equation} \\ 5((y^2+\frac{4}{5}y+\frac{4}{25})-\frac{4}{25})=0 \\ 5(y^2+\frac{4}{5}y+\frac{4}{25})-5(\frac{4}{25})=0 \\ 5(y^2+\frac{4}{5}y+\frac{4}{25})-\frac{4}{5}=0 \\ \\ (y^2+\frac{4}{5}y+\frac{4}{25})=(y+\frac{2}{5})^2 \\ \\ \therefore5(y+\frac{2}{5})^2-\frac{4}{5}=0 \\ \\ \text{ In standard form, the Equation becomes} \\ 5(y+\frac{2}{5})^2=\frac{4}{5} \\ \\ \\ \text{Step 4: Solve for }y \\ 5(y+\frac{2}{5})^2=\frac{4}{5} \\ \text{ Divide both sides by 5} \\ \frac{5}{5}(y+\frac{2}{5})^2=\frac{4}{5}\times\frac{1}{5} \\ (y+\frac{2}{5})^2=\frac{4}{25} \\ \\ \text{ Find the square root of both sides} \\ \sqrt[]{(y+\frac{2}{5})^2}=\pm\sqrt[]{\frac{4}{25}} \\ \\ y+\frac{2}{5}=\pm\frac{2}{5} \\ \\ \text{Subtract }\frac{2}{5}\text{ from both sides} \\ \\ y=-\frac{2}{5}\pm\frac{2}{5} \\ \\ \therefore y=0\text{ or }-\frac{4}{5} \end{gathered}[/tex]Final Answer
The solutions to the quadratic equations are
[tex]\begin{gathered} a^2-4a-45 \\ \text{Solution: }a=-5\text{ or }9 \\ \\ 5y^2+4y=0 \\ \text{Solution: }y=0\text{ or }-\frac{4}{5} \end{gathered}[/tex]