## Finding the roots of a quadratic in vertex form

If you have taken algebra 2, you know you can write quadratic functions in three forms:

• Standard form: $y = ax^2 + bx +c$
• Vertex form: $y = a(x-h)^2 + k$
• Factored form: $y = a(x-r_1)(x-r_2)$

No matter which of the forms you have, you are often asked to find the roots (x-intercepts). The factored form is already done for you: the roots are the values of $r_1$ and $r_2$. If your equation is in standard form, you use the quadratic formula. But what do you do if you have a quadratic in vertex form? Most students are taught to expand the equation into standard form (and then use the quadratic formula). But there’s a quick shortcut that is pretty easy to use. In fact, you can write the roots by inspection. If $0 = a(x-h)^2 + k$, then the roots are

$x = h \pm \sqrt{\dfrac{-k}{a}}$

There are three cases:

• $k=0$: There is one (double) root;
• $a \text{ and } k$ have different signs: the two roots are real;
• $a \text{ and } k$ have the same sign: the two roots are complex.

In the complex case, you can express the roots as

$x = h \pm \sqrt{\dfrac{k}{a}} \; i$

Here are two quick examples:

Example 1: $0 = 2(x-3)^2 + 6$

Using the formula above, the roots can be written immediately as

$x = 3 \pm \sqrt{\dfrac{6}{2}} \; i =3 \pm \sqrt{3} \; i$

Example 2: $0 = 2(x-3)^2 - 8$

The roots are:

$x = 3 \pm \sqrt{\dfrac{8}{2}} = 3 \pm 2 = 1 \text{ and } 5$

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