# The Quadratic Formula

## The Quadratic Formula Revision

**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.

**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 -

**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.

**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}}

**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**.

**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}

**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**).

## The Quadratic Formula Example Questions

**Question 1:** Solve 2x^2 - 4x = 1, giving your answers to 2 decimal places.

**[2 marks]**

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)}

**Question 2:** Find the discriminant of 4x - x^2 - 4 and determine how many roots 4x - x^2 - 4 = 0 has.

**[2 marks]**

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.

**Question 3:** f(x) = 4x^2 + x + 2k. Find the values of k for which f(x)=0 has two distinct real roots.

**[2 marks]**

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}

**Question 4:** (m+1)x^2 + (m+1)x + 2 = 0 has two distinct real solutions for x, with constant m.

**a)** Show that m^2 - 6m - 7 > 0

**b)** Hence, find the range for the possible values for m.

**[6 marks]**

**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