7.2 CHAPTER 7. SOLVINGQUADRATIC EQUATIONS
We see that sv =− 12 and s + 2v =− 5. This is a set of simultaneous equations
in s and v, but it is easy to solvenumerically. All the options for s and v are
considered below.
s v s + 2v
2 − 6 − 10
− 2 6 10
3 − 4 − 5
− 3 4 5
4 − 3 − 2
− 4 3 2
6 − 2 2
− 6 2 − 2
We see that the combination s = 3 and v =− 4 gives s + 2v =− 5.
Step 4 : Write the equation with factors
(2x + 3)(x− 4) = 0
Step 5 : Solve the equation
If two brackets are multiplied together and give0, then one of the brackets must
be 0, therefore
2 x + 3 = 0
or
x− 4 = 0
Therefore, x =−^32 or x = 4
Step 6 : Write the final answer
The solutions to 2 x^2 − 5 x− 12 = 0 are x =−^32 or x = 4.
It is important to remember that a quadratic equation has to be in the form ax^2 + bx + c = 0 before
one can solve it using the factorisation method.
Example 2: Solving quadratic equation by factorisation
QUESTION
Solve for a: a(a− 3) = 10
SOLUTION
Step 1 : Rewrite the equation inthe form ax^2 + bx + c = 0
Remove the brackets and move all terms to oneside.