Solutions of Quadratic Equations and Inequalities

There are 3 main methods that are used to solve quadratics.

Methods of Solving Quadratics

Factoring

This method is best applied by strong students who can inspect quadratics effectively. When a quadratic is in the form

$$0=ax^2 + bx + c$$

We can apply some smart tricks to factor this expression into two linear factors.

Let us suppose that the roots are $r_1, r_2$.

So we know that

$$0=a(x-r_1)(x-r_2)$$

$$0=a(x^2 - (r_1 + r_2)x + r_1r_2)$$

$$0=ax^2 - a(r_1 + r_2)x + ar_1r_2$$

Since we know $c$ and $b$, we look at the value of $c$ and the value of $b$ to try to find $r_1,r_2$ by factoring $c$ into two numbers that add to $b$

Completing the Square

The method of completing the square takes a quadratic and turns it into a perfect square trinomial, which is in the form

$$(x+k)^2 = x^2 + 2kx + k^2$$

This method turns a quadratic from standard form into vertex form.

$$y = ax^2 + bx + c$$

To do this method, we must make sure that the coefficient of $x^2 $ is 1, so we divide by $a$.

$$\frac{y}{a}= x^2 + \frac{b}{a}x + \frac{c}{a}$$

Now for the confusing part—for a perfect square trinomial, notice that the constant term is the coefficient of the $x$ term halved then squared.

$$(\frac{2k}{2})^2 = k^2$$

Therefore, to obtain a perfect square trinomial for $y$, we can add look at the coefficient of the $x$ term, and add a constant which has the same value as when said coefficient is halved then squared

$$\frac{y}{a} + \frac{b^2}{4a}= x^2 + \frac{b}{a}x + \frac{b^2}{4a}+ \frac{c}{a}$$

So now we can write the right hand side as

$$\frac{y}{a} + \frac{b^2}{4a^2}= (x+\frac{b}{2a})^2 + \frac{c}{a}$$

This allows us to isolate $x$, transitioning greatly into the next part

The Quadratic Formula

The quadratic formula a tool for solving any quadratic equation in the form $ax^2 + bx + c = 0$: Continuing on from completing the square, we have factored a quadratic equation from standard form to.

$$\frac{y}{a} + \frac{b^2}{4a^2}= (x+\frac{b}{2a})^2 + \frac{c}{a}$$

To solve a quadratic, we want to find its roots, so we can set $y=0$

$$ \frac{b^2}{4a^2}= (x+\frac{b}{2a})^2 + \frac{c}{a}$$

Doing some simple algebra:

$$ \frac{b^2}{4a^2}- \frac{c}{a}= (x+\frac{b}{2a})^2 $$

$$ (x+\frac{b}{2a})^2= \frac{b^2-4ac}{4a^2} $$

Now, by taking the square root (remember the $\pm$ sign)

$$ (x+\frac{b}{2a})= \pm\frac{\sqrt{b^2-4ac}}{2a} $$

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

The Discriminant

The discriminant ($\Delta$) is the expression under the square root in the quadratic formula:

$$ \Delta = b^2 - 4ac $$

The discriminant tells us everything about the nature of the roots:

• If $\Delta > 0$: Two distinct real roots (because then the $\pm$ sign would cause two values for $x$
• If $\Delta = 0$: One repeated real root (because $+0=-0$, so \pm sign does nothing in this case)
• If $\Delta< 0$: Two complex conjugate roots (Because you can't take the square root of a negative number without introducing imaginary numbers)

Solving Quadratic Inequalities

When solving quadratic inequalities like $ax^2 + bx + c > 0$ or $ax^2 + bx + c<0$:

1. Find the roots of the corresponding equation $ax^2 + bx + c = 0$
2. Plot these points on a number line
3. Test regions between points to determine where inequality holds

Method for Inequalities:

Draw the quadratic out, paying attention to its concavity, and look at when it is $<0$ or $>0$ with respect to the roots.

The rest is obvious.