1.05 Solving quadratic equations

Quadratic equations

The three equations below are considered quadratic equations due to the highest index of $2$2 on any x term.

$x^2=25$x2=25

$2x^2-4x=4$2x2−4x=4

$x^2+5x-6=0$x2+5x−6=0

There are a number of ways to solve quadratic equations. The solutions to quadratic equations are also called the roots of the equation - we will cover this language in more detail in a later section.

Solving quadratic equations

Methods

The four methods we have seen are:

• Algebraically solve - for simple binomial quadratics such as $x^2=49$x2=49 which have no $x$x term, we can rearrange and solve using a square root. Remember that when we take a square root we need to consider both the positive and negative root. For example:

$x^2$x2	$=$=	$49$49
$x$x	$=$=	$\pm7$±7

• Factorise - fully factorising a quadratic allows us to use the null factor law: if $a\times b=0$a×b=0 then at least one of $a=0$a=0 or $b=0$b=0 must be true. For example:

$x^2-2x-3$x2−2x−3	$=$=	$0$0
$(x+1)(x-3)$(x+1)(x−3)	$=$=	$0$0
$x$x	$=$=	$-1,3$−1,3

• Completing the square - this method involves rewriting the equation such that there is only one $x$x present. We can then continue to solve the equation algebraically. We will review this method in more detail later in the lesson.
• Quadratic formula - using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a​ will find the solutions to any quadratic equation of the form $ax^2+bx+c=0$ax2+bx+c=0. The quadratic formula is useful when we can't algebraically solve, find a nice integer factorisation of the quadratic or completing the square introduces fractions into the working.

Strategy

When looking to solve a quadratic, we should start by checking the simplest options:

• Is there an $x$x term in the quadratic? If not, we can solve it straight away algebraically
• Are there any common factors that we can take out to simplify the quadratic?
• Can we find an easy factorisation with nice integers?
• If not, consider completing the square or using the quadratic formula
   

Practice questions

Question 1

Question 2

Question 3

Completing the square

If the quadratic is not easily factorisable, then we can either complete the square or use the quadratic formula.

The video in the following example demonstrates one method for completing the square.

Worked example

Question 1

Solve the quadratic equation $x^2+4x-5=0$x2+4x−5=0 by first completing the square.

Think: The following video shows how we can complete the square for the expression $x^2+4x-5$x2+4x−5.

Do: We can now solve this algebraically.

$x^2+4x-5$x2+4x−5	$=$=	$0$0
$\left(x+2\right)^2-9$(x+2)2−9	$=$=	$0$0
$\left(x+2\right)^2$(x+2)2	$=$=	$9$9
$x+2$x+2	$=$=	$\pm3$±3
$x$x	$=$=	$3-2,-3-2$3−2,−3−2
$x$x	$=$=	$1,-5$1,−5

 

We can also complete the square using an algebraic method. The steps involved in this are:

• Move the constant term to the other side of the equation
• Take half of the coefficient of $x$x, square it, and add the result to both sides to keep the equation balanced
• Complete the square

For this example:

$x^2+4x-5$x2+4x−5	$=$=	$0$0
$x^2+4x$x2+4x	$=$=	$5$5
$x^2+4x+4$x2+4x+4	$=$=	$5+4$5+4
$\left(x+2\right)^2$(x+2)2	$=$=	$9$9
$x+2$x+2	$=$=	$3-2,-3-2$3−2,−3−2
$x+2$x+2	$=$=	$\pm3$±3
$x$x	$=$=	$3-2,-3-2$3−2,−3−2
$x$x	$=$=	$1,-5$1,−5

Other quadratic equations

We might come across equations with fractions that don't look like quadratic equations at first glance. For example, the equations $x-4=-\frac{6}{x}$x−4=−6x​ and $\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}$1x​−1x+1​=12​ don't immediately look like quadratic equations. If we multiply each term by the lowest common denominator, however, we can see the true quadratic nature of the equation. For example, if we multiply both sides of $x-4=-\frac{6}{x}$x−4=−6x​ by $x$x we get:

$x^2-4x$x2−4x	$=$=	$-6$−6
$x^2-4x+6$x2−4x+6	$=$=	$0$0

 

We can solve this using one of our current methods.

Practice questions

Question 4

Question 5

Question 6