1.2.6 Solving quadratic equations

What you'll learn

How to solve quadratic equations by factorising, completing the square, and using the quadratic formula.
How the discriminant tells you how many real roots a quadratic has.
How to recognise and solve equations that are quadratic in a function of the unknown.
How to check solutions and choose an efficient method.

Recognising a quadratic equation

A quadratic equation can be written in the form

ax^2+bx+c=0,

where $a$ , $b$ , and $c$ are constants and $a≠0a\neq 0$ .

The highest power of the unknown is 2. Before applying a standard solving method, rearrange the equation so that one side is zero.

Definition

Roots

The roots, or solutions, of an equation are the values of the unknown that make the equation true. On the graph $y=ax^2+bx+c$ , real roots are the $x$ -coordinates where the curve meets the $x$ -axis.

For example, $x^2=5x-6$ should first be rearranged to

x^2-5x+6=0.

A quadratic may have two distinct real roots, one repeated real root, or no real roots.

Three quadratic graphs showing two real roots, one repeated root, and no real roots

Solving by factorising

Factorising is usually the quickest method when the quadratic splits into simple linear factors.

This method uses the zero-product property: if $pq = 0$ , then $p = 0$ or $q = 0$ .

Key Idea

Factorise only when one side is zero

You can conclude that either factor is zero only when their product equals zero. Rearrange the equation before factorising.

Example

Solving a quadratic by factorising

Solve $2x^2-7x+3=0$ .

To factorise, find two terms whose product is $2x2⋅3=6x22x^2\cdot 3=6x^2$ and whose sum is $- 7 x$ . These are $- 6 x$ and $- x$ :
$2x^2-6x-x+3=0.$
Factorise by grouping:
$2x(x−3)−1(x−3)=0,(2x−1)(x−3)=0.\begin{aligned} 2x(x-3)-1(x-3)&=0,\\ (2x-1)(x-3)&=0. \end{aligned}$
Apply the zero-product property:
$\quad \text{or} \quad x-3=0.$
Solve each linear equation:
$x=12orx=3.x=\frac12 \quad \text{or} \quad x=3.$

Common Mistake

Losing one of the roots

A quadratic can have two solutions. After factorising, set each linear factor equal to zero.

Solving by completing the square

To complete the square means rewriting a quadratic using a squared bracket. For a monic quadratic, whose coefficient of $x^2$ is 1,

x2+bx=(x+b2)2−(b2)2.x^2+bx=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2.

Completing the square is useful when factorisation is not obvious. It also reveals the turning point of the corresponding quadratic graph.

Example

Solving by completing the square

Solve $x^2+6x+1=0$ , giving exact answers.

Take half the coefficient of $x$ , which is 3, and complete the square:
$x^2+6x+1=(x+3)^2-9+1=(x+3)^2-8.$
Set the completed-square form equal to zero and isolate the square:
$(x+3)2−8=0,(x+3)2=8.\begin{aligned} (x+3)^2-8&=0,\\ (x+3)^2&=8. \end{aligned}$
Take both the positive and negative square roots:
$x+3=±8=±22.x+3=\pm\sqrt8=\pm 2\sqrt2.$
Subtract 3:
$x=−3±22.x=-3\pm 2\sqrt2.$

Common Mistake

Forgetting the negative square root

From $u^2=k$ , you need $u=±ku=\pm\sqrt{k}$ when $k > 0$ . Using only the positive square root loses a solution.

The quadratic formula

Every quadratic equation $ax^2+bx+c=0$ , where $a≠0a\neq 0$ , can be solved using the quadratic formula:

x=−b±b2−4ac2a.x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

This is especially useful when the quadratic does not factorise easily.

Tip

Substitute coefficients with their signs

Write down $a$ , $b$ , and $c$ before substituting. If a coefficient is negative, keep it in brackets during the calculation.

Example

Using the quadratic formula

Solve $3x^2+2x-7=0$ , giving exact answers.

Identify the coefficients:
$a=3,b=2,c=−7.a=3,\qquad b=2,\qquad c=-7.$
Substitute them into the formula:
$x=−2±22−4(3)(−7)2(3).x=\frac{-2\pm\sqrt{2^2-4(3)(-7)}}{2(3)}.$
Simplify the expression under the square root:
$2^2-4(3)(-7)=4+84=88.$
Simplify the surd and the resulting fraction:
$x=−2±886=−2±2226=−1±223.\begin{aligned} x&=\frac{-2\pm\sqrt{88}}6\\ &=\frac{-2\pm 2\sqrt{22}}6\\ &=\frac{-1\pm\sqrt{22}}3. \end{aligned}$

Unless a decimal answer is requested, leave irrational solutions in exact surd form.

The discriminant and the number of roots

The expression inside the square root in the quadratic formula is called the discriminant.

Definition

Discriminant

For $ax^2+bx+c=0$ , the discriminant is

Δ=b2−4ac.\Delta=b^2-4ac.

Its sign determines the number of real roots.

If $Δ>0\Delta>0$ , there are two distinct real roots.
If $Δ=0\Delta=0$ , there is one repeated real root.
If $Δ<0\Delta<0$ , there are no real roots.

A repeated root occurs when both solutions have the same value. Graphically, the parabola touches the $x$ -axis at its turning point.

Example

Determining the number of real roots

Determine the number of real roots of $x^2-4x+7=0$ .

Identify $a = 1$ , $b = - 4$ , and $c = 7$ .
Calculate the discriminant:
$Δ=(−4)2−4(1)(7)=16−28=−12.\begin{aligned} \Delta&=(-4)^2-4(1)(7)\\ &=16-28\\ &=-12. \end{aligned}$
Since $Δ<0\Delta<0$ , the equation has no real roots. The corresponding graph does not meet the $x$ -axis.

Common Mistake

Squaring a negative coefficient

If $b = - 4$ , then $b^2=(-4)^2=16$ , not $- 16$ . Use brackets when substituting a negative value.

Quadratics in a function of the unknown

Some equations do not initially look quadratic because the repeated expression is more complicated than $x$ . They may still have the structure

A[f(x)]^2+Bf(x)+C=0.

This is called an equation quadratic in a function of the unknown. Temporarily substitute $u = f (x)$ , solve the quadratic in $u$ , and then return to the original unknown.

Common disguised quadratics include equations involving $x^4$ and $x^2$ , or two powers such as $3^{2x}$ and $3^x$ .

Key Idea

Spot the repeated expression

Look for one expression and its square. For example, $x^4$ is the square of $x^2$ , while $x+1)^4$ is the square of $x+1)^2$ .

Example

Solving a disguised quadratic

Solve

x+1)^4-5(x+1)^2+4=0.

The equation is quadratic in $x+1)^2$ . Let
$u=(x+1)^2.$
Then $x+1)^4=u^2$ , so the equation becomes
$u^2-5u+4=0.$
Factorise and solve the quadratic in $u$ :
$(u−1)(u−4)=0,u=1oru=4.\begin{aligned} (u-1)(u-4)&=0,\\ u&=1 \quad \text{or} \quad u=4. \end{aligned}$
Replace $u$ with $x+1)^2$ :
$(x+1)2=1or(x+1)2=4.(x+1)^2=1 \quad \text{or} \quad (x+1)^2=4.$
Solve both resulting equations, including positive and negative square roots:
$x=−3.\begin{aligned} x+1&=\pm 1 &&\Rightarrow x=0\text{ or }x=-2,\\ x+1&=\pm 2 &&\Rightarrow x=1\text{ or }x=-3. \end{aligned}$
Therefore, the four solutions are
$1.x=-3,\,-2,\,0,\,1.$

Common Mistake

Check the range of your substitution

A substituted variable may have restrictions. If $u=x^2$ , then $u≥0u\geq 0$ , so any negative solution for $u$ must be rejected before returning to $x$ .

Checking your solutions

You can check an exact solution by substituting it into the original equation. For decimal solutions, substitution may give a value close to zero because of rounding.

A graphing calculator can also provide a useful check: the calculated roots should match the $x$ -intercepts. However, your written solution must still show the required algebra.

Exam technique

In the exam

Rearrange the equation into the form $ax^2+bx+c=0$ before choosing a method.
Try factorising first when integer factors look likely; otherwise use completing the square or the quadratic formula.
For a disguised quadratic, state your substitution clearly and remember to solve back in terms of the original unknown.
Keep exact values unless the question requests decimals, and check that you have included every possible root.

Self review

Check yourself

Can you solve $6x^2-x-2=0$ by factorising?
What does $b^2-4ac=0$ tell you about the roots and the graph?
How would you begin solving $x^6-7x^3+12=0$ ?

What you'll learn

How to solve quadratic equations by factorising, completing the square, and using the quadratic formula.
How the discriminant tells you how many real roots a quadratic has.
How to recognise and solve equations that are quadratic in a function of the unknown.
How to check solutions and choose an efficient method.

Recognising a quadratic equation

A quadratic equation can be written in the form

ax^2+bx+c=0,

where $a$ , $b$ , and $c$ are constants and $a≠0a\neq 0$ .

The highest power of the unknown is 2. Before applying a standard solving method, rearrange the equation so that one side is zero.

Definition

Roots

For example, $x^2=5x-6$ should first be rearranged to

x^2-5x+6=0.

A quadratic may have two distinct real roots, one repeated real root, or no real roots.

Three quadratic graphs showing two real roots, one repeated root, and no real roots

Solving by factorising

Factorising is usually the quickest method when the quadratic splits into simple linear factors.

This method uses the zero-product property: if $pq = 0$ , then $p = 0$ or $q = 0$ .

Key Idea

Factorise only when one side is zero

You can conclude that either factor is zero only when their product equals zero. Rearrange the equation before factorising.

Example

Solving a quadratic by factorising

Solve $2x^2-7x+3=0$ .

To factorise, find two terms whose product is $2x2⋅3=6x22x^2\cdot 3=6x^2$ and whose sum is $- 7 x$ . These are $- 6 x$ and $- x$ :
$2x^2-6x-x+3=0.$
Factorise by grouping:
$2x(x−3)−1(x−3)=0,(2x−1)(x−3)=0.\begin{aligned} 2x(x-3)-1(x-3)&=0,\\ (2x-1)(x-3)&=0. \end{aligned}$
Apply the zero-product property:
$\quad \text{or} \quad x-3=0.$
Solve each linear equation:
$x=12orx=3.x=\frac12 \quad \text{or} \quad x=3.$

Common Mistake

Losing one of the roots

A quadratic can have two solutions. After factorising, set each linear factor equal to zero.

Solving by completing the square

To complete the square means rewriting a quadratic using a squared bracket. For a monic quadratic, whose coefficient of $x^2$ is 1,

x2+bx=(x+b2)2−(b2)2.x^2+bx=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2.

Completing the square is useful when factorisation is not obvious. It also reveals the turning point of the corresponding quadratic graph.

Example

Solving by completing the square

Solve $x^2+6x+1=0$ , giving exact answers.

Take half the coefficient of $x$ , which is 3, and complete the square:
$x^2+6x+1=(x+3)^2-9+1=(x+3)^2-8.$
Set the completed-square form equal to zero and isolate the square:
$(x+3)2−8=0,(x+3)2=8.\begin{aligned} (x+3)^2-8&=0,\\ (x+3)^2&=8. \end{aligned}$
Take both the positive and negative square roots:
$x+3=±8=±22.x+3=\pm\sqrt8=\pm 2\sqrt2.$
Subtract 3:
$x=−3±22.x=-3\pm 2\sqrt2.$

Common Mistake

Forgetting the negative square root

From $u^2=k$ , you need $u=±ku=\pm\sqrt{k}$ when $k > 0$ . Using only the positive square root loses a solution.

The quadratic formula

Every quadratic equation $ax^2+bx+c=0$ , where $a≠0a\neq 0$ , can be solved using the quadratic formula:

x=−b±b2−4ac2a.x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

This is especially useful when the quadratic does not factorise easily.

Tip

Substitute coefficients with their signs

Write down $a$ , $b$ , and $c$ before substituting. If a coefficient is negative, keep it in brackets during the calculation.

Example

Using the quadratic formula

Solve $3x^2+2x-7=0$ , giving exact answers.

Identify the coefficients:
$a=3,b=2,c=−7.a=3,\qquad b=2,\qquad c=-7.$
Substitute them into the formula:
$x=−2±22−4(3)(−7)2(3).x=\frac{-2\pm\sqrt{2^2-4(3)(-7)}}{2(3)}.$
Simplify the expression under the square root:
$2^2-4(3)(-7)=4+84=88.$
Simplify the surd and the resulting fraction:
$x=−2±886=−2±2226=−1±223.\begin{aligned} x&=\frac{-2\pm\sqrt{88}}6\\ &=\frac{-2\pm 2\sqrt{22}}6\\ &=\frac{-1\pm\sqrt{22}}3. \end{aligned}$

Unless a decimal answer is requested, leave irrational solutions in exact surd form.

The discriminant and the number of roots

The expression inside the square root in the quadratic formula is called the discriminant.

Definition

Discriminant

For $ax^2+bx+c=0$ , the discriminant is

Δ=b2−4ac.\Delta=b^2-4ac.

Its sign determines the number of real roots.

If $Δ>0\Delta>0$ , there are two distinct real roots.
If $Δ=0\Delta=0$ , there is one repeated real root.
If $Δ<0\Delta<0$ , there are no real roots.

A repeated root occurs when both solutions have the same value. Graphically, the parabola touches the $x$ -axis at its turning point.

Example

Determining the number of real roots

Determine the number of real roots of $x^2-4x+7=0$ .

Identify $a = 1$ , $b = - 4$ , and $c = 7$ .
Calculate the discriminant:
$Δ=(−4)2−4(1)(7)=16−28=−12.\begin{aligned} \Delta&=(-4)^2-4(1)(7)\\ &=16-28\\ &=-12. \end{aligned}$
Since $Δ<0\Delta<0$ , the equation has no real roots. The corresponding graph does not meet the $x$ -axis.

Common Mistake

Squaring a negative coefficient

If $b = - 4$ , then $b^2=(-4)^2=16$ , not $- 16$ . Use brackets when substituting a negative value.

Quadratics in a function of the unknown

Some equations do not initially look quadratic because the repeated expression is more complicated than $x$ . They may still have the structure

A[f(x)]^2+Bf(x)+C=0.

This is called an equation quadratic in a function of the unknown. Temporarily substitute $u = f (x)$ , solve the quadratic in $u$ , and then return to the original unknown.

Common disguised quadratics include equations involving $x^4$ and $x^2$ , or two powers such as $3^{2x}$ and $3^x$ .

Key Idea

Spot the repeated expression

Look for one expression and its square. For example, $x^4$ is the square of $x^2$ , while $x+1)^4$ is the square of $x+1)^2$ .

Example

Solving a disguised quadratic

Solve

x+1)^4-5(x+1)^2+4=0.

The equation is quadratic in $x+1)^2$ . Let
$u=(x+1)^2.$
Then $x+1)^4=u^2$ , so the equation becomes
$u^2-5u+4=0.$
Factorise and solve the quadratic in $u$ :
$(u−1)(u−4)=0,u=1oru=4.\begin{aligned} (u-1)(u-4)&=0,\\ u&=1 \quad \text{or} \quad u=4. \end{aligned}$
Replace $u$ with $x+1)^2$ :
$(x+1)2=1or(x+1)2=4.(x+1)^2=1 \quad \text{or} \quad (x+1)^2=4.$
Solve both resulting equations, including positive and negative square roots:
$x=−3.\begin{aligned} x+1&=\pm 1 &&\Rightarrow x=0\text{ or }x=-2,\\ x+1&=\pm 2 &&\Rightarrow x=1\text{ or }x=-3. \end{aligned}$
Therefore, the four solutions are
$1.x=-3,\,-2,\,0,\,1.$

Common Mistake

Check the range of your substitution

A substituted variable may have restrictions. If $u=x^2$ , then $u≥0u\geq 0$ , so any negative solution for $u$ must be rejected before returning to $x$ .

Checking your solutions

You can check an exact solution by substituting it into the original equation. For decimal solutions, substitution may give a value close to zero because of rounding.

A graphing calculator can also provide a useful check: the calculated roots should match the $x$ -intercepts. However, your written solution must still show the required algebra.

Exam technique

In the exam

Rearrange the equation into the form $ax^2+bx+c=0$ before choosing a method.
Try factorising first when integer factors look likely; otherwise use completing the square or the quadratic formula.
For a disguised quadratic, state your substitution clearly and remember to solve back in terms of the original unknown.
Keep exact values unless the question requests decimals, and check that you have included every possible root.

Self review

Check yourself

Can you solve $6x^2-x-2=0$ by factorising?
What does $b^2-4ac=0$ tell you about the roots and the graph?
How would you begin solving $x^6-7x^3+12=0$ ?

What you'll learn

Recognising a quadratic equation

Solving by factorising

Solving by completing the square

The quadratic formula

The discriminant and the number of roots

Quadratics in a function of the unknown

Checking your solutions

Review 1.2.6 Solving quadratic equations by teaching Genie

1.2.6 Solving quadratic equations

What you'll learn

Recognising a quadratic equation

Solving by factorising

Solving by completing the square

The quadratic formula

The discriminant and the number of roots

Quadratics in a function of the unknown

Checking your solutions

Review 1.2.6 Solving quadratic equations by teaching Genie

1.2.6 Solving quadratic equations

What you'll learn

Recognising a quadratic equation

Solving by factorising

Solving by completing the square

The quadratic formula

The discriminant and the number of roots

Quadratics in a function of the unknown

Checking your solutions

Review 1.2.6 Solving quadratic equations by teaching Genie

1.2.6 Solving quadratic equations Revision Guide

1.2.6 Solving quadratic equations

What you'll learn

Recognising a quadratic equation

Solving by factorising

Solving by completing the square

The quadratic formula

The discriminant and the number of roots

Quadratics in a function of the unknown

Checking your solutions

Review 1.2.6 Solving quadratic equations by teaching Genie

1.2.6 Solving quadratic equations Revision Guide