What are the possible value(s) of x for the following: x^2 + 3x - 54 = 0

It is a quadratic equation and when factorised will be of the form (x +/- a) (x +/- b) = 0. We need to find 2 numbers, a and b, that add together to make the 3 for the 3x term and which are multiplied together to give -54 for the constant term. Because the constant is negative (-54), one of the numbers multiplied to give -54 must be negative and the other must be positive as a positive x negative = negative. Possible numbers:

1.(+ or -)1 x (+ or -)54 = -54

2.(+ or -)2 x (+ or -)27 = -54

3.(+ or -)3 x (+ or -)18 = -54

4.(+ or -)6 x (+ or -)9 = -54

We can exclude the first 3 possibilities as it wouldn't be possible to add them together to make +3 (for 3x). This leaves us with (+ or -)6 and (+ or -)9. If we had (-9) + (+6) we would get -3. If we had (+9) + (-6) we would get +3 which is correct.

This means that a= +9 and b= -6

(x + 9) (x - 6) = 0

In order for the above to =0, one of the brackets needs to=0 because any number, n x 0 = 0

Either: x + 9 = 0 which is rearranged to give x= -9 or x - 6 = 0 which is rearranged to give x= 6.

Therefore x= 6 or -9