TutorChase logo
CIE A-Level Maths Study Notes

1.1.3 Solving Quadratic Equations

In this section, we'll explore various techniques for solving quadratic equations. These include factorisation, completing the square, and the quadratic formula. Additionally, we'll examine how to handle equations in quadratic form, which, although not inherently quadratic, can be transformed into quadratic equations.

Solving Equations in Quadratic Form

Some equations, while not quadratic initially, can be transformed into quadratic equations. This is achieved by substituting a function of xx into the equation.

Example 1:

Solve x45x2+4=0x^4 - 5x^2 + 4 = 0.

Solution:

Let u=x2u = x^2, then the equation becomes u25u+4=0u^2 - 5u + 4 = 0. Solving for uugives:

(u4)(u1)=0u=4,1(u - 4)(u - 1) = 0 \Rightarrow u = 4, 1

Thus, x=±ux=±2,±1x = \pm \sqrt{u} \Rightarrow x = \pm 2, \pm 1.

Example 2:

Solve 6x+x1=06x + \sqrt{x} - 1 = 0.

Solution:

Let u=xu = \sqrt{x}, transforming the equation to 6u2+u1=06u^2 + u - 1 = 0. Solving for uu gives:

(3u1)(2u+1)=0u=13,12(3u - 1)(2u + 1) = 0 \Rightarrow u = \frac{1}{3}, -\frac{1}{2}

Since x=u2x = u^2, we reject u=12u = -\frac{1}{2} as it yields no real solutions. Therefore, x=(13)2=19x = \left(\frac{1}{3}\right)^2 = \frac{1}{9}.

Factorising Quadratic Equations

Factorising is effective when the quadratic can be easily broken down into factors.

Example 1:

Solve x25x+6=0x^2 - 5x + 6 = 0.

Solution:

x25x+6=(x2)(x3)=0x=2,3x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \Rightarrow x = 2, 3

Example 2:

Solve x24x5=0x^2 - 4x - 5 = 0.

Solution:

x24x5=(x5)(x+1)=0x=5,1x^2 - 4x - 5 = (x - 5)(x + 1) = 0 \Rightarrow x = 5, -1

Completing the Square

This method is useful for quadratic equations that are not easily factorisable.

Example 1:

Solve x2+4x5=0x^2 + 4x - 5 = 0.

Solution:

x2+4x=5x^2 + 4x = 5(x+2)24=5(x + 2)^2 - 4 = 5 (x+2)2=9(x + 2)^2 = 9x+2=±3x + 2 = \pm 3 x=1,5x = 1, -5

Example 2:

Solve x26x+8=0x^2 - 6x + 8 = 0.

Solution:

x26x=8x^2 - 6x = -8(x3)29=8(x - 3)^2 - 9 = -8(x3)2=1(x - 3)^2 = 1x3=±1x - 3 = \pm 1 x=4,2x = 4, 2

Using the Quadratic Formula

This formula, x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, can solve any quadratic equation.

Example 1:

Solve 2x23x+1=02x^2 - 3x + 1 = 0.

Solution:

x=3±(3)242122x=1,12x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \Rightarrow x = 1, \frac{1}{2}

Example 2:

Solve x2x2=0x^2 - x - 2 = 0.

Solution:

x=1±1241(2)21x=2,1x = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \Rightarrow x = 2, -1

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2
Your details
Alternatively contact us via
WhatsApp, Phone Call, or Email