Question

Explain how you would use each method to solve the equation

$x^2-4x-5=0$

(a) By factoring

(b) By completing the square

(c) By using the Quadratic Formula

Short answer

(a)

$\begin{array}{l}{x}^2-4x-5=0\\ \Rightarrow \left(x+1\right)\left(x-5\right)=0\\ \Rightarrow x=-1\text{or}x=5\end{array}$

(b)

$\begin{array}{l}{x}^2-4x-5=0\\ \Rightarrow x^2-4x=5\\ \Rightarrow {\left(x-2\right)}^2=3^2\\ \Rightarrow {\left(x-2\right)}^2=\pm 3\\ \Rightarrow x=5\text{or}x=-1\end{array}$

(c)

$\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\left(1\right)(-5)}}{2\left(1\right)}\\ \Rightarrow x=5\text{or}x=-1\end{array}$

Step-by-step solution

Part a. Step 1. Apply the concept of factoring.

Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. If a quadratic equation can be factored, it is written as a product of linear terms.

Part a. Step 2. Analyze the given expression.

Given

$x^2-4x-5=0$

This is a quadratic equation.

The middle term here is$-4x$ that can be factored as$-4x=-5x+x$

Part a. Step 3. Solve.

Solve the given expression by factoring the middle term:

$\begin{array}{l}{x}^2-4x-5=0\\ \Rightarrow x^2-5x+x-5=0\\ \Rightarrow x\left(x-5\right)+\left(x-5\right)=0\\ \Rightarrow \left(x+1\right)\left(x-5\right)=\mathrm{0...............}.....\left[\text{DistributiveProperty}\right]\\ \Rightarrow x+1=0\text{or}x-5=0\\ \Rightarrow x=-1\text{or}x=5\end{array}$

Part b. Step 1. Apply the completing the square.

Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.

To solve $a{x}^2+bx+c=0$by completing the square:

1. Transform the equation so that the constant term, c, is alone on the right side.
2. If a, the leading coefficient (the coefficient of the $x^2$term), is not equal to 1. Divide both sides by a.
3. Add the square of half the coefficient of the x -term, ${\left(\frac{b}{2a}\right)}^2$to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Take the square root of both sides. (Remember: ${\left(x+q\right)}^2=r$is equivalent to $x+q=\pm \sqrt{r}$)
6. Solve for x.

Part b. Step 2. Analyze the given expression.

Given

$x^2-4x-5=0$

This is a quadratic equation.

Part b. Step 3. Solve.

Solve the given expression by completing the square:

$\begin{array}{l}{x}^2-4x-5=0\\ \Rightarrow x^2-4x=5\\ \Rightarrow x^2-4x+{\left(\frac{4}{2}\right)}^2=5+{\left(\frac{4}{2}\right)}^2\\ \Rightarrow x^2-4x+2^2=5+2^2\\ \Rightarrow x^2-2\cdot 2\cdot x+2^2=9\\ \Rightarrow {\left(x-2\right)}^2=3^2\\ \Rightarrow {\left(x-2\right)}^2=\pm 3\\ \Rightarrow x-2=3\text{or}x-2=-3\\ \Rightarrow x=5\text{or}x=-1\end{array}$

Part c. Step 1. Apply the quadratic formula.

Quadratic Formula is the simplest way to find the roots of a quadratic equation.

For $a{x}^2+bx+c=0$, the roots are given by the quadratic formula$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Part c. Step 2. Analyze the given expression.

Given

$x^2-4x-5=0$

This is a quadratic equation.

Here $a=1$, $b=-4$, $c=-5$

Part c. Step 3. Solve.

Roots of the equation by quadratic formula:

$\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\left(1\right)(-5)}}{2\left(1\right)}\\ \Rightarrow x=\frac{4\pm \sqrt{16+20}}{2}\\ \Rightarrow x=\frac{4\pm \sqrt{36}}{2}\\ \Rightarrow x=\frac{4\pm 6}{2}\\ \Rightarrow x=\frac{4+6}{2}\text{or}x=\frac{4-6}{2}\\ \Rightarrow x=\frac{10}{2}\text{or}x=\frac{-2}{2}\\ \Rightarrow x=5\text{or}x=-1\end{array}$