Question

How do you find the real solutions of the polynomial `2x^3=19x^2-49x+20`?

Answer 1

The roots are `1/2`, `5` and `4`.

Explanation:

Given:

`2x^3=19x^2-49x+20`

Subtract the right hand side from the left to get the standard form:

`2x^3-19x^2+49x-20 = 0`

By the rational root theorem, any rational zeros of this cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-20` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational roots are:

`+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-10, +-20`

We find that `x = 1/2` is a zero...

`2(color(blue)(1/2))^3-19(color(blue)(1/2))^2-49(color(blue)(1/2))+20 = (1-19-98+80)/4 = 0`

Hence `(2x-1)` is a factor:

`2x^3-19x^2+49x-20 = (2x-1)(x^2-9x+20)`

To factor the remaining quadratic, find a pair of factors of `20` with sum `9`. The pair `5, 4` works and hence:

`x^2-9x+20 = (x-5)(x-4)`

So the other two roots are `x=5` and `x=4`.