Question

Suppose that r1 and r2 are the roots of the quadratic equation ax2 + bx + c = 0. Note that: (x – r1)(x – r2)= x2 – (r1 + r2)x + r1r2. Explain how to use this result to check the solutions of a quadratic equation of the form x2 + bx + c. Then use the result to check that –3 + 2i and –3 – 2i are solutions of the equation x2 + 6x + 13 =0.

Answer 1

Thank you for asking this question. Here is your answer:

The equation must be  

(x -r1)(x - r2) = 0 (If r1 and r2 are roots of as monic quadratic)

Now we will expand the equation:

x² -(r1 + r2) + r1r2 = 0 = x² + bx + c

We will divide the equation by the coefficient which is leading, or you can divide it by a.

We will get this:

ax² + bx + c as a(x - r1)(x - r2) = 0  

or ax² -a(r1 + r2)x + ar1r2 = 0

If there is any confusion please leave a comment below.