How to find the roots in a quadratic function in vertex form?

Question

I have been assigned the task to express the vertex form quadratic function from 2(x - (sqrt(2)/2))^2 - 3 - sqrt(2) into the standard form and the x-intercept form. The vertex form, in my reference, is f(x) = a(x-h)^2 + k. How can I convert this into the standard form f(x) = ax^2 + bx + c and from there find the roots and find the root form f(x) = a(x-r)(x-s) where r and s are roots?

Answer 1

While you can get it into standard form by using FOIL or the binomial theorem to square $(x-h)^2=x^2-2hx+h^2$, it's actually easier not to go through the standard form:

$$\begin{align} a(x-h)^2+k&=0 \\ a(x-h)^2&=-k \\ (x-h)^2&=-k/a \\ x-h&=\pm\sqrt{-k/a}\\ x&=h\pm \sqrt{-k/a} \end{align}.$$

Answer 2

To convert to "standard" form, expand and simplify.

To convert to "$x$-intercept" form, factorise as follows. We have $$2(x-\sqrt 2/2)^2-3-\sqrt 2=2(x-\sqrt 2/2)^2-\left(\sqrt{3+\sqrt2}\right)^2,$$ which is a difference of two squares and may be factorised by noting that $$x^2-y^2=(x-y)(x+y).$$