Question

How do you solve `9x^2-5x=7`?

Answer 1

Subtract `7` from both sides to get: `9x^2-5x-7 = 0`, then solve using the quadratic formula to get:

`x = (5+-sqrt(277))/18`

Explanation:

`9x^2-5x-7` is of the form `ax^2+bx+c`, with `a=9`, `b=-5` and `c=-7`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (-5)^2-(4xx9xx-7) = 25+252`

`= 277`

Since `Delta > 0`, the quadratic equation has two distinct Real roots. Since `Delta` is not a perfect square (`277` is prime), those roots are irrational.

The roots are given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`=(-b+-sqrt(Delta))/(2a)`

`=(5+-sqrt(277))/18`

Notice the discriminant `Delta` is the expression under the square root.

So if `Delta < 0` the square root is not Real and the quadratic has no Real roots - It has a conjugate pair of distinct complex roots.

If `Delta = 0` then there is one repeated Real root.

If `Delta > 0` (as in our example), there are two distinct Real roots. If in addition `Delta` is a perfect square, then those roots are rational.