Solving quadratic equations

​​In a nutshell

A quadratic equation can be given in the form $$ax^2+bx+c=0$$ where $$a$$, $$b$$ and $$c$$ are constants (and $$a\neq0$$). Quadratic equations can be solved using factorising or by using the quadratic formula. These methods can also help you identify the number of solutions to a quadratic equation.

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Quadratic equations

Quadratic equations can have two, one or zero solutions. The methods below will find all of the solutions.

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Factorising

A brief reminder of factorising is given below:

PROCEDURE

Factorising allows you to solve a quadratic equation since once a quadratic is written in the form $$(Ax+C)(Bx+D)$$, you have:

$$(Ax+C)(Bx+D)=0$$​​

Notice that you are multiplying two terms together: $$Ax+C$$ and $$Bx+D$$. Since their product is zero, it follows that at least one of the terms equals zero itself. Since you don't know which one, you can suppose either of them are in turn:

​$$Ax+C=0$$​​

​$$x=\frac{-C}{A}$$​​

or

​$$Bx+D=0$$​​

​$$x=\frac{-D}{B}$$

Example 1

Solve by factorising the following quadratic.

$$x^2-10x=-21$$​​

The first step is to express this equation in the form $$ax^2+bx+c=0$$:

$$x^2-10x+21=0$$​​

Now you can factorise:

$$(x-3)(x-7)=0$$​​

So either $$x-3=0$$:

$$\underline{x=3}$$​​

or $$x-7=0$$:

$$\underline{x=7}$$​​

Hence they both satisfy the equation $$x^2-10x=-21$$ and are thus both solutions.

The quadratic formula

The quadratic formula is

​$$\boxed{x= \dfrac{-b \pm \sqrt{b^2-4ac}} {2a}}$$​​

Where $$a,b$$ and $$c$$ are the coefficients from the quadratic equation in the form

$$ax^2+bx+c=0$$

This directly solves a quadratic equation. Notice that the $$\pm$$ is how you can have two solutions: use the plus for one solution and use the minus for the other. Sometimes, these solutions will be the same anyway.

Example 2

By using the quadratic formula, solve the equation $$6x^2-5x-4=0$$.

It is already in the correct form, so start by identifying $$a$$, $$b$$ and $$c$$:

$$a=6$$, $$b=-5$$ and $$c=-4$$

Note: It is essential that, if it is part of the coefficient, you include the negative sign when identifying these three values.

Now insert these values directly into the formula before using the calculator:

$$x= \dfrac{-(-5) \pm \sqrt{(-5)^2-4(6)(-4)}} {2(6)}$$​​

You don't have to use a calculator yet - you can simplify first:

$$x= \dfrac{5 \pm \sqrt{121}} {12}= \dfrac{5 \pm 11} {12}$$​​

Sometimes, you may not need a calculator at all. Here you have that 

$$x= \dfrac{5 +11} {12}=\dfrac{16}{12}=\dfrac{4}{3}$$​​

by using the plus, and

$$x= \dfrac{5 -11} {12}=\dfrac{-6}{12}=-\dfrac{1}{2}$$​​

Hence the solutions are $$\underline{x=\dfrac43}$$ and $$\underline{x=-\dfrac12}$$.

Solve quadratics using a calculator

Quadratic equations can be solved using the equation function. Select Equation/Func, polynomial, degree $$2$$, and then enter $$a, b$$ and $$c$$.

CALCULATOR TIP

​$$\boxed{\begin{aligned}&Equation/Func \\&Polynomial \\&Degree: 2 \\&ax^2+bx+c \\&a: \qquad b: \qquad c: \qquad\end{aligned}}$$​​

Example 3

Use your calculator to solve the equation

​$$2x^2-9x-5=0$$​​

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Solve the quadratic by identifying $$a=2, b=-9$$ and $$c=-5$$.

	​$$\boxed{\begin{aligned}&Equation/Func \\&Polynomial \\&Degree: 2 \\&ax^2+bx+c \\&a:2 \quad b:-9 \quad c:-5 \quad\end{aligned}}$$​​
	​$$\boxed{x_1= 5 \ and \ x_2=-\frac 1 2 \quad}$$​​

Therefore, $$\underline{x_1=5}$$ and $$\underline{x_2=-\frac 1 2 }$$​​