Solving Quadratic Equations (AQA GCSE Further Maths): Revision Note

Solving quadratic equations

How do I solve a quadratic equation using factorisation?

• Factorise the quadratic and solve each bracket equal to zero

• To solve $\left(2x-3\right)\left(3x+5\right)=0$

  • solve 2x – 3 = 0 to get x = $\frac{3}{2}$

  • solve 3x + 5 = 0 to get x = $-\frac{5}{3}$

• To solve $x\left(x-4\right)=0$ don't forget to solve x = 0

  • The two solutions are x = 0 or x = 4

    • It is a common mistake to divide by x at the beginning (you will lose a solution)

How do I solve a quadratic equation by completing the square?

• To solve x² + bx + c = 0 

  • replace the first two terms, x² + bx, with (x + p)² - p² where p is half of b

  • this is called completing the square

    • x² + bx + c = 0 becomes

  • rearrange this equation to make x the subject (using ±√)

• For example, solve x² + 10x + 9 = 0 by completing the square

  • x² + 10x becomes (x + 5)² - 5²

  • so x² + 10x + 9 = 0 becomes (x + 5)² - 5² + 9 = 0

  • make x the subject (using ±√)
    

    • (x + 5)² - 25 + 9 = 0

    • (x + 5)² = 16

    • x + 5 = ±√16

    • x  = ±4 - 5

    • x  = -1 or x  = -9

• If the equation is ax² + bx + c = 0 with a number in front of x², then divide both sides by a first, before completing the square 

How do I use the quadratic formula to solve a quadratic equation?

• The quadratic formula is 

  • $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

• Read off the values of a, b and c from the equation

• Substitute these into the formula

  • Write this line of working in the exam

  • Put brackets around any negative numbers being substituted in

• To solve 2x² - 7x - 3 = 0 using the quadratic formula:

  • a = 2, b = -7 and c = -3

  • $x=\frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\times 2\times \left(-3\right)}}{2\times 2}$

  • Type this into a calculator

    • once with + for  ± and once with - for  ±

  • The solutions are x = 3.886 and x = -0.386 (to 3 dp)

    • Rounding is often asked for in the question

• The calculator also gives these solutions in exact form (surd form)

  • x = $\frac{7+\sqrt{73}}{4}$ and x = $\frac{7-\sqrt{73}}{4}$

  • You need to be able to find solutions in exact / surd form without a calculator

    • this means working out (-7)² - 4 × 2 × (-3)

What is the discriminant?

• The part of the formula under the square root (b² – 4ac) is called the discriminant

• The sign of this value tells you if there are 0, 1 or 2 solutions

  • If b² – 4ac > 0 (positive)

    • then there are 2 different solutions

  • If b² – 4ac = 0 (zero) 

    • then there is only 1 solution

    • sometimes called "repeated solutions"

  • If b² – 4ac < 0 (negative)

    • then there are no solutions

    • If your calculator gives you solutions with i terms in, these are "complex" and not what we are looking for

  • Interestingly, if b² – 4ac is a perfect square number ( 1, 4, 9, 16, …) then the quadratic expression could have been factorised!

How does completing the square link to the quadratic formula?

• The quadratic formula actually comes from completing the square of ax² + bx + c = 0

• You can see hints of this when you solve quadratics 

  • For example, solving x² + 10x + 9 = 0 

    • by completing the square, (x + 5)² = 16 so x  = -5 ± 4

    • by the quadratic formula,  $x=\frac{-10\pm \sqrt{64}}{2}=-5\pm \frac{8}{2}$ = -5 ± 4

Can I use my calculator to solve quadratic equations?

• Yes, in the calculator paper, use a calculator to check your final solutions!

  • Calculators also help you to factorise (if you're struggling with that step)

• A calculator gives solutions to $6x^2+x-2=0$ as x = $-\frac{2}{3}$  and x = $\frac{1}{2}$

  • "Reverse" the method above to factorise

    • $6x^2+x-2\equiv \left(3x+2\right)\left(2x-1\right)$

  • Warning: a calculator gives solutions to 12x² + 2x – 4 = 0 as x = $-\frac{2}{3}$ and x = $\frac{1}{2}$ 

    • But 12x² + 2x – 4 $\not\equiv \left(3x+2\right)\left(2x-1\right)$

    • the right-hand side expands to 6x² + ... ,not 12x² + ...

    • Correct this by multiplying the right by 2

    • 12x² + 2x – 4 $\equiv 2\left(3x+2\right)\left(2x-1\right)$

Hidden quadratic equations

How do I spot a hidden quadratic equation?

• Hidden quadratics have the same structure as quadratic equations

  • a(something)² + b(something) + c = 0

• Here are some hidden quadratics based on x² - 3x - 4 = 0:

  • $x^4-3x^2-4=0$ (a quadratic in x²)

  • $x^{16}-3x^8-4=0$ (a quadratic in x⁸)

  • $x-3\sqrt{x}-4=0$ (a quadratic in $\sqrt{x}$ because ${\left(\sqrt{x}\right)}^2$ is $x$)

  • $x^{\frac{2}{3}}-3x^{\frac{1}{3}}-4=0$ (a quadratic in $x^{\frac{1}{3}}$because ${\left(x^{\frac{1}{3}}\right)}^2=x^{\frac{2}{3}}$)

• Sometimes, a change of base helps to spot a hidden quadratic

  • e.g. the first term in $4^x-3\times 2^x-4=0$ can be written $4^x={\left(2^2\right)}^x=2^{2x}={\left(2^x\right)}^2$

    • ${\left(2^x\right)}^2-3\times 2^x-4=0$ is a quadratic in $2^x$

How do I solve a hidden quadratic equation?

• You can solve a(...)² + b(...) + c = 0 with a substitution

  • Substitute "u = ..." and rewrite the equation in terms of u only 

    • au² + bu + c = 0

  • Solve this easier quadratic equation in u to get u = p and u = q

  • Replace the u's with their substitution to get two equations

    • "... = p" and "... = q"

  • Solve these two separate equations to find all the solutions

    • These equations might have multiple solutions or none at all!

• e.g. to solve x⁴ - 3x² - 4 = 0

  • Substitute u = x² to get u² - 3u - 4 = 0

  • The solutions are u = 4 or u = -1,

  • Rewrite in terms of x: 

    • x² = 4 or x² = -1,

  • Solve to give x = -2 or x = 2 (no solutions from x² = -1 as you can't square-root a negative)