Question

How do you write a quadratic equation in standard form with solutions 1 and -8?

Answer 1

The roots of a polynomial are, in a certain sense, the blocks to build the polynomial itself.

Take for example an expression like `f(x)=(x-\alpha)`, for some real number `\alpha`. This is a line (first degree polynomial) which equals zero exacly when `x=\alpha`, since `f(\alpha) = \alpha-\alpha=0`.

To build a polynomial with higher degree, simply multiply such blocks: if we consider `g(x) = (x-\alpha)(x-\beta)`, this function will have exactly `\alpha` and `\beta` as roots, since now we have

`g(\alpha) = (\alpha-\alpha)(\alpha-\beta) = 0*(\alpha-\beta) = 0`
`g(\beta) = (\beta-\alpha)(beta-\beta) = (\beta-\alpha)*0 = 0`

In this way, you can build a polynomial of the degree you want, with the roots that you want. In your case, you have to build

`p(x) = (x-1)(x+8)`

In fact,

`p(1) = (1-1)(1+8) = 0*9 = 0`
`p(-8) = (-8-1)(-8+8) = -9*0 = 0`

So, `p(x)` has the roots we want it to have. To write it in standard form, simply expand the parenthesis:

`(x-1)(x+8) = x^2+7x-8`

Faster Strategy:

In the particular case of quadratic polynomials, we have a faster way to solve exercises like this: the generic expression

`x^2+ax+b` (note that the `x^2` coefficient must be `1`)

yields two solutions `x_1` and `x_2` such that

`x_1+x_2 = -a`
`x_1x_2 = b`

In other words, the sum of the solutions is the opposite of the `x` coefficient, and the product of the solutions is the last coefficient of the quadratic expression.

If your solutions have to be `1` and `-8`, then we have

`a=-(1-8) = 7`
`b = 1*(-8) = -8`

and again we come to the conclusion that

`p(x) = x^2+7x-8`