1.1.3 Solving Quadratic Equations

In this topic we will learn how to:

solve quadratic equations, by factorising, completing the square and using the formula

There are three ways of solving quadratic equations:

$\textbf{\textcolor{gray}{Factorisation}}$ A quadratic equation can be solved by sight using factorization.

1. When the coefficient of $x^{2}$ is $1$ .

To factorise the quadratic $x^{2} + bx + c$ , find the pair of factors of $c$ that add up to $b$ .
Let’s take a look at the example below.

Example 1
Solve $x^{2} + 5x - 6 = 0$ by factorization.
$x^{2} + \textcolor{#0f0}{5}x\ \textcolor{red}{-\ 6} = 0$ The first step is to open two sets of parentheses next to each other each with an $x$ in them and equate them to $0$ ,
$(x\ \ \ \ )(x\ \ \ \ ) = 0$ Identify $\textcolor{#0f0}{b}$ and $\textcolor{red}{c}$ ,
$\textcolor{#0f0}{b = 5}\ \ \ \ \ \ \textcolor{red}{c = -6}$ Find pair factors of $\textcolor{red}{c}$ ,

$-6$ and $1$ ; $6$ and $-1$ ; $3$ and $-2$ ; $-2$ and $3$

Find the pair of factors that add up to $\textcolor{#0f0}{b}$ ,
$6 \textmd{ and } -1$ Add each number to one of the two parentheses we opened earlier,
$(x + 6)(x - 1) = 0$ Note: If you were to expand the two sets of parentheses you should get the original equation.

Equate each bracket to $0$ ,
$x + 6 = 0\ \ \ \ \ \ x - 1 = 0$ Solve the two linear equations,
$x = -6\ \ \ \ \ \ x = 1$ Therefore, the roots are,
$x = -6,\ \ x = 1$ 2. When the coefficient of $x^{2}$ is not $1$

Factorizing a quadratic equation when the coefficient of $x^{2}$ is not $1$ is a bit more challenging, however, with enough practice, it will become easier. Let’s take a look at the example below.

Example 2
Solve $2x^{2} + 9x + 10 = 0$ by factorization.
$\textcolor{#2192ff}{2}x^{2} + \textcolor{#0f0}{9}x + \textcolor{red}{10} = 0$ The first step is to identify $a$ , $b$ and $c$ ,
$\textcolor{#2192ff}{a = 2},\ \ \textcolor{#0f0}{b = 9},\ \ \textcolor{red}{c = 10}$ Then find the product of $\textcolor{#2192ff}{a}$ and $\textcolor{red}{c}$ ,
$\textcolor{#2192ff}{2} \times \textcolor{red}{10} = 20$ List the pair factors of $20$ ,
$20\textmd{ and }1\textmd{; }10\textmd{ and }2\textmd{; }5\textmd{ and }4\textmd{; }-20\textmd{ and }-1\textmd{; }\\ -10\textmd{ and }-2\textmd{; }-5\textmd{ and }-4$ Find the pair of factors that add up to $\textcolor{#0f0}{b}$ ,
$5 \textmd{ and } 4$ Rewrite $\textcolor{#0f0}{b}$ as the sum of these factors,
$2x^2 + \textcolor{#0f0}{9}x + 10 = 0$ $2x^2 + \textcolor{#0f0}{(5 + 4)}x + 10 = 0$ Remove the parentheses,
$2x^2 + \textcolor{#0f0}{5x + 4x} + 10 = 0$ Now we will factorize by grouping. Group terms that are multiples of each other,
$(2x^2 + 4x) + (5x + 10) = 0$ Factorise the expressions in parentheses,
$2x\textcolor{#2192ff}{(x + 2)} + 5\textcolor{#2192ff}{(x + 2)} = 0$ We then factor out the $\textcolor{#2192ff}{(x + 2)}$ since it is common,
$(2x + 5)(x + 2) = 0$ Equate each bracket to $0$ ,
$2x + 5 = 0\ \ \ \ \ \ x + 2 = 0$ Solve the two linear equations,
$x = -\frac{5}{2}\ \ \ \ \ \ x = -2$ Therefore, the roots are,
$x = -\frac{5}{2},\ \ x = -2$ Note: With practice, most of the steps outlined above will become intuitive and you can skip them. Factorising by sight should be the method you use in solving a quadratic equation unless told otherwise.

$\textbf{\textcolor{gray}{Completing the square}}$ To solve a quadratic equation using the completing the square method you first have to complete the square. After completing the square, make $x$ the subject of the formula.
Let’s take a look at an example below.

Example 3
Solve $x^{2} + 5x - 6 = 0$ by first completing the square.
$x^{2} + 5x - 6 = 0$ The first step is to complete the square,
$\left(x + \frac{5}{2}\right)^{2} - \left(\frac{5}{2}\right)^{2} - 6 = 0$ Simplify,
$\left(x + \frac{5}{2}\right)^{2} \textcolor{#2192ff}{- \frac{49}{4}} = 0$ Once you have completed the square you have to make $x$ the subject of the formula. To do that we will start by moving the term outside the parentheses to the other side of the equal sign,
$\left(x + \frac{5}{2}\right)^{2} = \textcolor{#2192ff}{\frac{49}{4}}$ Take the square root of both sides to get rid of the power 2,
$\sqrt{\left(x + \frac{5}{2}\right)^{2}} = \pm\sqrt{\frac{49}{4}}$ Note: We put a $\pm$ sign, whenever we take the square root of a number.

The square root gets rid of the power $2$ on the left hand side,
$x + \frac{5}{2} = \pm\frac{7}{2}$ Make $x$ the subject of the formula,
$x =-\frac{5}{2} \pm\frac{7}{2}$ Since there is a $\pm$ sign, we can split the equation above into two separate equations,
$x = -\frac{5}{2} + \frac{7}{2}\ \ \ \ \ \ x = -\frac{5}{2} - \frac{7}{2}$ So our equation has two solutions which are,
$x = 1\ \ \ \ \ \ x = -6$

$\textbf{\textcolor{gray}{Quadratic Formula}}$ For the quadratic equation,
$ax^{2} + bx + c = 0$ The quadratic formula is,
$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ This can be used to solve any quadratic equation, where $a$ , $b$ and $c$ are known constants.
Let’s take a look at an example below.

Example 4
Solve $x^{2} + 5x - 6 = 0$ using the quadratic formula.

The first step is to identify the values of $a$ , $b$ , and $c$ ,
$\textcolor{#2192ff}{a = 1},\ \ \textcolor{#0f0}{b = 5},\ \ \textcolor{red}{c = -6}$ Substitute $\textcolor{#2192ff}{a = 1}$ , $\textcolor{#0f0}{b = 5}$ , $\textcolor{red}{c = -6}$ into the quadratic formula,
$x = \frac{\textcolor{#0f0}{-5} \pm \sqrt{(\textcolor{#0f0}{-5})^{2} - 4(\textcolor{#2192ff}{1})(\textcolor{red}{-6})}}{2(\textcolor{#2192ff}{1})}$ Simplify,
$x = \frac{-5 \pm \sqrt{25 + 24}}{2}$ $x = \frac{-5 \pm \sqrt{49}}{2}$ $x = \frac{-5 \pm 7}{2}$ The above can be written as two separate equations,
$x = -\frac{5}{2} + \frac{7}{2}\ \ \ \ \ \ x = -\frac{5}{2} - \frac{7}{2}$ So our final solutions are,
$x = 1,\ \ x = -6$

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