The Quadratic Formula

The Quadratic Formula

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

The Quadratic Formula

The Quadratic Formula is used to find the solutions to any quadratic equation.

The following topics are referred to in this page.

A Level AQA Edexcel OCR

The Quadratic Formula

The solutions to the quadratic equation

\textcolor{red}{a} x^2 + \textcolor{blue}{b}x + \textcolor{limegreen}{c} = 0

are given by the quadratic formula:

x=\dfrac{-\textcolor{blue}{b}\pm\sqrt{\textcolor{blue}{b}^2-4\textcolor{red}{a}\textcolor{limegreen}{c}}}{2\textcolor{red}{a}}

Note: There are two solutions for x: one using + and the other using -

A LevelAQAEdexcelOCR

The Discriminant

The discriminant of the quadratic formula (written as D or \Delta) is the part under the square root sign. i.e. the discriminant is

b^2-4ac

It can be positive, zero or negative – these tell us how many roots the quadratic equation has:

  • If b^2 - 4ac > 0, then the quadratic has 2 real roots (that are distinct)
  • If b^2 - 4ac = 0, then the quadratic has 1 real root (or ‘equal roots’)
  • If b^2 - 4ac < 0, then the quadratic has no real roots

This can be seen visually:

Note: We say no ‘real’ roots since some quadratics can have ‘imaginary’ roots – however we will not see this in this course.

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Using the Quadratic Formula

Find the solutions to the quadratic equation 3x^2 - 8x + 2=0, giving your answers in surd form.

[2 marks]

a = 3, b = -8 and c = 2

Put these values into the quadratic formula:

\begin{aligned} x &= \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \\ &= \dfrac{8 \pm \sqrt{40}}{6} \\ &= \dfrac{4 \pm \sqrt{10}}{3} \end{aligned}

 

So, x = \textcolor{orange}{\dfrac{4 + \sqrt{10}}{3}} or \textcolor{orange}{x = \dfrac{4 - \sqrt{10}}{3}}

A Level AQA Edexcel OCR

Example 2: Finding the Discriminant

How many real roots does the quadratic equation 2x^2 - 5x + 3=0 have?

[2 marks]

a = 2, b = -5 and c = 3

So, the discriminant is

b^2-4ac = (-5)^2 - 4 \times 2 \times 3 = 1

The discriminant is >0, so 2x^2 - 5x + 3=0 has two real roots.

A Level AQA Edexcel OCR

Example 3: Using the Discriminant

f(x) = 2x^2 + 4x + k. Find the values of k for which f(x)=0 has no real roots.

[3 marks]

a = 2, b = 4 and c = k

The discriminant is

b^2 - 4ac = 4^2 - 4 \times 2 \times k = 16 - 8k

The quadratic equation has no real roots, therefore b^2 - 4ac < 0

So,

\begin{aligned} 16 - 8k &< 0 \\ 16 &< 8k \\ \textcolor{orange}{k} &\textcolor{orange}{> 2} \end{aligned}

A Level AQA Edexcel OCR

Example 4: Using the Discriminant

kx^2 + 2kx + 3 = 0 has two distinct real roots.

Find the set of values for which k satisfies this.

[5 marks]

a = k, b = 2k and c = 3

The discriminant is

b^2 - 4ac = (2k)^2 - (4 \times k \times 3) = 4k^2 - 12k

Since the quadratic equation has two distinct real roots, the discriminant must be >0

4k^2 - 12k > 0

The quadratic can be factorised:

4k^2 - 12k = 4k(k-3)

This is 0 when k=0 or when k=3

Hence, 4k(k-3)>0 when \textcolor{orange}{k<0} or when \textcolor{orange}{k>3}

 

Note: You can draw a graph of y = 4k(k-3) to help – you would see a u-shaped graph that crosses the k-axis at 0 and 3 (see inequalities).

A Level AQA Edexcel OCR

Example Questions

Rearrange the equation so that it is in the form ax^2 + bx + c = 0:

 

2x^2 - 4x - 1 = 0

 

Then, a = 2, b = -4 and c = -1

 

Put these values into the quadratic formula:

 

\begin{aligned} x &= \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 2 \times -1}}{2 \times 2} \\ &= \dfrac{4 \pm \sqrt{24}}{4} \\ &= \dfrac{2 \pm \sqrt{6}}{2} \end{aligned}

So,

x = \dfrac{2 + \sqrt{6}}{2} = 2.22 \text{  (2 dp)}

x = \dfrac{2 - \sqrt{6}}{2} = -0.22 \text{  (2 dp)}

Rearrange the equation so that it is in the form ax^2 + bx + c = 0:

 

-x^2 + 4x - 4 = 0

So, a = -1, b = 4 and c = -4

 

Then, find the discriminant:

b^2 - 4ac = 4^2 - 4(-1)(-4) = 0

 

Hence, 4x - x^2 - 4 = 0 has one real distinct root.

a = 4, b=1 and c = 2k

 

Find the discriminant:

b^2 - 4ac = 1^2 - 4 \times 4 \times 2k = 1 - 32k

 

f(x)=0 has two distinct real roots, therefore b^2-4ac > 0:

 

\begin{aligned} 1 - 32k &> 0 \\ 1 &> 32k \\ k &< \dfrac{1}{32} \end{aligned}

a) a = (m+1), b =(m+1) and c = 2

 

Find the discriminant:

\begin{aligned} b^2 - 4ac &= (m+1)^2 - 4 \times (m+1) \times 2 \\ &= (m^2 + 2m + 1) - (8m + 8) \\ &= m^2 -6m - 7 \end{aligned}

 

The equation has two real distinct solutions, therefore b^2 - 4ac >0:

 

m^2 -6m - 7 > 0

 

b) m^2 -6m - 7 = (m-7)(m+1)

 

This expression is 0 when k = 7 and k = -1

 

Hence, (m-7)(m+1)>0 when m< -1 or when m>7

Related Topics

MME

Inequalities

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

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