Find the roots of the quadratic equation below:
$$z^{2} - 14z + 53$$

Show Solution

The way to solve this problem is to use the quadratic formula, that you'll be familiar with from GCSE Maths, remember this is:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
for a quadratic equation of the form:
$$ax^{2} + bx + c = 0$$
This is the same form that we have our equation in already (except we have ??z?? instead of ??x??) - so we have no need to rearrange and can start
substituting values in.

The values of ??a??, ??b?? and ??c?? can be found easily in this case:
$$a = 1$$
$$b = -14$$
$$c = 53$$

And substituting into the quadratic equation gives the answer:
$$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$z = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \times 1 \times 53}}{2 \times 1} $$
$$z = \frac{14 \pm \sqrt{196 - 212}}{2}$$
$$z = \frac{14 \pm \sqrt{-16}}{2}$$
Now, we can deal with the negative square root by splitting it up:
$$\sqrt{-16} = \sqrt{16 \times -1} = \sqrt{16} \times \sqrt{-1} = 4 \times \sqrt{-1} = 4i$$
Substituting this back into the quadratic equation gives the final answer:
$$z = \frac{14 \pm 4i}{2}$$
$$z = 7 \pm 2i$$

This question is from Edexcel - Further Pure 1 - Complex Numbers, why not try more questions
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