A brief note on Solving Quadratic Equations by Factorising . Read it before a test. Try each question at the end before you open its answer.
Quadratic equations Standard form: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 , with a ≠ 0 a \neq 0 a = 0 . The numbers b b b and c c c may be 0 0 0 . Expand and collect every term on one side before you decide. In ( x + 4 ) 2 = x 2 + 2 x (x + 4)^2 = x^2 + 2x ( x + 4 ) 2 = x 2 + 2 x the x 2 x^2 x 2 terms cancel, so it is linear. A root makes the left side 0 0 0 . Test it by putting it in. A quadratic equation has at most two roots, where its curve meets the x x x axis. Solving by factorisation If a product is 0 0 0 , one of its factors is 0 0 0 . Split the middle term: find p p p and q q q with p + q = b p + q = b p + q = b and p × q = a × c p \times q = a \times c p × q = a × c , write b x = p x + q x bx = px + qx b x = p x + q x , and group. Each factor gives a root: ( 2 x − 3 ) = 0 (2x - 3) = 0 ( 2 x − 3 ) = 0 gives x = 3 2 x = \dfrac{3}{2} x = 2 3 . Special cases: no constant term, take out x x x ; a difference of squares; a perfect square gives a repeated root. Word problems Name the unknown with its unit, write the equation, solve, and check the meaning. Keep only the roots that fit: lengths, speeds, ages and counts are not negative. Time = = = distance ÷ \div ÷ speed. Clear fractions by multiplying by every denominator. A worked example Solve 3 x 2 − 5 x − 2 = 0 3x^2 - 5x - 2 = 0 3 x 2 − 5 x − 2 = 0 .
The target. a × c = − 6 a \times c = -6 a × c = − 6 and b = − 5 b = -5 b = − 5 . The pair. − 6 -6 − 6 and 1 1 1 : product − 6 -6 − 6 , sum − 5 -5 − 5 . Split and group. 3 x 2 − 6 x + x − 2 = 3 x ( x − 2 ) + ( x − 2 ) = ( 3 x + 1 ) ( x − 2 ) 3x^2 - 6x + x - 2 = 3x(x - 2) + (x - 2) = (3x + 1)(x - 2) 3 x 2 − 6 x + x − 2 = 3 x ( x − 2 ) + ( x − 2 ) = ( 3 x + 1 ) ( x − 2 ) . Roots. x = 2 x = 2 x = 2 or x = − 1 3 x = -\dfrac{1}{3} x = − 3 1 . Check: 12 − 10 − 2 = 0 12 - 10 - 2 = 0 12 − 10 − 2 = 0 . Practice with new numbers
A program to try The program finds the split for the worked example and checks both roots.
Run it. Is the left side 0 for both roots?
1. Change the equation to a, b, c = 2, 7, 3. What split and roots does it find?
Mistakes to avoid Deciding too early. Simplify before you call an equation quadratic. Using c c c instead of a × c a \times c a × c when you search for the split. Wrong sign in a root. ( x + 4 ) = 0 (x + 4) = 0 ( x + 4 ) = 0 gives x = − 4 x = -4 x = − 4 . Dividing by x x x . It loses the root x = 0 x = 0 x = 0 . Keeping an impossible root in a word problem. Check yourself Solve x 2 − 8 x + 15 = 0 x^2 - 8x + 15 = 0 x 2 − 8 x + 15 = 0 . Stuck? Show a hint Show the answerIs x ( x + 1 ) = x 2 + 4 x(x + 1) = x^2 + 4 x ( x + 1 ) = x 2 + 4 a quadratic equation? Stuck? Show a hint Show the answerA rectangle is 2 m 2\ \text{m} 2 m longer than it is wide, and its area is 48 m 2 48\ \text{m}^2 48 m 2 . Find its width. Stuck? Show a hint Show the answerWhat are all the roots of x 2 = 3 x x^2 = 3x x 2 = 3 x ? Stuck? Show a hint Show the answer