Quadratic Formula (SQA National 5 Maths): Revision Note

Solving a quadratic equation using the quadratic formula

What is the quadratic formula?

• A quadratic equation has the form ax² + bx + c = 0 (where a ≠ 0)

  • you need "= 0" on one side

• The quadratic formula is a formula that gives both solutions to a quadratic equation:

$x=\frac{-b\pm \sqrt{\left(b^2-4ac\right)}}{2a}$

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

• 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 - 8x - 3 = 0 using the quadratic formula:

  • a = 2, b = -8 and c = -3

  • $x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\times 2\times \left(-3\right)}}{2\times 2}$

  • Either type this into a calculator or simplify by hand

    • Type it once using + for  ± then again using - for  ±

  • The solutions are x = 4.345207... or x = -0.345207....

    • To 3 decimal places: x = 4.345 or x = -0.345

    • To 3 significant figures: x = 4.35 or x = -0.345

How do I write the solutions in an exact (surd) form?

• It is also possible to give answers in an exact (surd) form

• For the example above, work out the number under the square root sign

  • Be careful with negatives!

    • ${\left(-8\right)}^2-4\times 2\times \left(-3\right)=64+24=88$

  • Now take the square root of this number and use surd rules to simplify

    • $\sqrt{88}=\sqrt{4\times 22}=\sqrt{4}\times \sqrt{22}=2\sqrt{22}$

  • Substitute this back into the formula and simplify

    • $x=\frac{8\pm 2\sqrt{22}}{4}=\frac{2\left(4\pm \sqrt{22}\right)}{4}=\frac{4\pm \sqrt{22}}{2}$

    • The solutions in exact (surd) form are $x=\frac{4+\sqrt{22}}{2}$ or $x=\frac{4-\sqrt{22}}{2}$

• Calculators that can solve quadratics will give solutions in exact (surd) form

Can I use my calculator to solve quadratic equations?

• If your calculator solves quadratic equations, use it to check your final answers

  • But a correct method and working must still be shown

• If your calculator gives you solutions with terms containing $\mathbf{i}$, these are "complex" and are not what you are looking for

  • You may have made a mistake entering the quadratic equation

  • Or the equation may not have any real number solutions