## Big Picture

Graphing and factoring are just some of the ways to solve quadratic equations. Graphing would not be a very accurate way to solve quadratic equations if the answers are not whole number integers, and quadratic equations cannot always be factored. There are, however, a number of other ways to solve quadratic equations, such as finding square roots and completing the square.

## Key Terms

Quadratic Equation: A nonlinear equation that can be written as $$ax2+bx+c = 0$$ where $$a \ne 0$$.

Perfect Square: A number with a whole-number square root.

Trinomial: A polynomial with three terms.

Quadratic Formula: For $$ax^2+bx+c = 0$$, the quadratic formula gives $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

Root (of a polynomial): The value that makes the polynomial equal to 0.

Discriminant: The expression inside the square root of the quadratic formula: $$b^2-4ac.$$

## Solving by Square Roots

In special cases where $$x^2+c = 0$$, we can solve the quadratic equation by taking the square root.

• First rearrange the equation so that $$x2$$ is the only term on one side of the equation.
• Then take the square root.
For $$x^2 = d$$,

• If $$d > 0$$, there are two solutions:  $$x = {\pm \sqrt{d} }.$$
• If $$d = 0$$, then there is one solution: $$x = 0$$.
• If $$d < 0$$, there are no solutions: we can’t take the square root of a negative number.

## Completing the Square

For any expression that looks like $$x^2 + bx$$, we can always add a constant c so that $$x^2+bx+c$$ is a perfect square trinomial. Perfect square trinomial:

• $$(x+a)^2 = x^2+2ax+a^2$$
• $$(x-a)^2 = x^2-2ax+a^2$$

To complete the square for $$x^2 + bx$$, add $$\Big(\frac {b} {2} \Big)^2$$.

• $$x^2 + bx + \Big(\frac {b} {2} \Big)^2 = \Big(x + \frac {b} {2} \Big)^2$$

### Solving Quadratic Equations by Completing the Square

• Start with $$ax^2+bx+c = 0$$.
• Move c to the other side: $$ax^2+bx = -c$$.
• Divide by a: $$x^2 + \frac {b} {a} x = -\frac {c} {a}$$
• Add $$\Big(\frac {\frac{b}{a}} {2} \Big)^2$$ = $$\frac {b^2} {4a^2}$$ to both sides so that the left side is a perfect square: $$\big(x+\frac{b}{2a}\big)^2 = \frac{b^2}{4a^2} - \frac{c}{a}$$
• Take the square root of both sides of the equation: $$x + \frac{b}{2a} = \pm \sqrt{ \frac{b^2}{4a^2} - \frac{c}{a}}$$.
• Subtract $$\frac{b}{2a}$$  to find $$x$$: $$x = \pm \sqrt{ \frac{b^2}{4a^2} - \frac{c}{a}} - \frac {b}{2a}$$.
• Don’t forget to simplify radical expressions. Simplifying the expression will give us the quadratic formula
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

These are common mistakes: make sure to avoid them!

• Don’t forget to divide by a: $$x^2$$ needs to have a coeficient of $$1$$.
• Don’t forget to put $$\pm$$  when taking the square root.
• In step $$4$$, when completing the square, make sure you add the term to both sides.

## Vertex Form

Completing the square is useful for rewriting quadratic equations into the form: $$y-k = a(x-h)^2$$.

• When written in this form, the vertex is at $$(h, k)$$.
• This makes it easy to graph the parabola.

To rewrite $$ax^2+bx+c = y$$,

• Move c to the other side: $$ax^2+bx = y-c$$.
• Divide by a: $$x^2 + \frac {b} {a} x = \frac {y - c} {a}$$
• Add $$\Big(\frac {\frac{b}{a}} {2} \Big)^2$$ = $$\frac {b^2} {4a^2}$$ to both sides so that the left side is a perfect square: $$\big(x+\frac{b}{2a}\big)^2 = \frac{b^2}{4a^2} - \frac{y - a}{a}$$.
• Multiply by a: $$a\big(x + \frac {b}{2a}\big)^2 = y - c + \frac{b^2}{4a}$$.

So the vertex is $$\big(- \frac{b}{2a}, c - \frac{b^2}{4a}\big)$$

The quadratic formula comes directly from completing the square. By memorizing the quadratic formula, we do not have to follow the multi-step process of completing the square.

If we know $$ax^2+bx+c = 0$$, then we can find $$x$$.

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$.

• Make sure the quadratic equation equals $$0$$ before using the quadratic formula!
• This gives us the roots of the equation. Knowing the roots and the vertical line $$x= -\frac{b}{2a}$$, we can graph the parabola.

### Discriminant

The discriminant can tell us if a quadratic equation will have any solutions before we even begin to solve it. The discriminant is under the square root. We can only take the square root of numbers greater than or equal to $$0$$.

• $$b^2 - 4ac < 0$$ means taking the square root of a negative number, which means there are no solutions
• $$b^2 - 4ac = 0$$ means taking the square root of $$0$$, which means there is only one solution
• $$b^2 - 4ac > 0$$ means taking the square root of a positive number, which means there are two solutions