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The quadratic formula, which was first discovered by the Babylonians about four thousand years ago, gives you a surefire way to solve quadratic equations of the form:

$0=a{x}^{2}+bx+c$

These solutions are known as the zeros of x, and by substituting the known values of the coefficients a, b, and c, you can solve for the zeros of x. The quadratic formula looks like this:

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

Since quadratic equations have a degree equal to two, there will be up to two real solutions to the equation. This is what the (±) in the formula indicates.

Recall that a quadratic equation is unlike a linear equation in that it has an ${x}^{2}$ (x-squared) term. "Quadratum" is the Latin term for squared. Quadratic equations are polynomial equations of the second degree, which indicates that it has at least one term that is squared. Quadratic equations generally look like this:

$a{x}^{2}+bx+c=0$

where x is an unknown variable, and a, b, and c are numerical coefficients. For example, ${x}^{2}+3x+6$ is a quadratic equation. Here, a 0 because if $a=0$ then the equation will not remain quadratic and will instead be linear, such as:

$bx+c=0$

The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations. The roots of any polynomial are the solutions for the given equation.

## Standard form of the quadratic equation

A quadratic equation is an equation that can be written in the standard form:

$a{x}^{2}+bx+c=0$

where a, b, and c are real numbers and a does not equal 0

Because the quadratic equation contains only one variable. The power of x must be a non-negative integer. So, the equation is a polynomial where the highest power is 2.

The solutions for the quadratic equation are the zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratic equations, there are up to two real roots or zeros that satisfy the equation. If we put the values of roots or x on the left-hand side of the equation, it will equal zero. This is why they are termed zeros.

Using the quadratic formula is a guaranteed way to find the solution(s) to any quadratic equation in the form:

$0=a{x}^{2}+bx+c$

where

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

The expression under the square root sign $\left({b}^{2}–4ac\right)$ is called the discriminant. If this is negative, there are no real solutions to the equation. If the discriminant is zero, there is only one real solution. If the discriminant is positive, then the symbol indicates that you will get two real solutions.

Example 1

${x}^{2}–x–12=0$

Here, $a=1$ , $b=-1$ , and $c=-12$ . Substituting, we get

$x=-\left(-1\right)±\frac{\sqrt{\left({\left(-1\right)}^{2}-4\left(1\right)\left(-12\right)\right)}}{2\left(1\right)}$

Simplify.

$x=\frac{1±\sqrt{49}}{2}$

The discriminant is positive, so we have two solutions:

$x=\frac{8}{2}$ and $x=-\frac{6}{2}$

$x=4$ and $x=-3$

In this example, the discriminant was 49, a perfect square, so we ended up with rational answers. Often, when using the quadratic formula, you end up with answers that still contain radicals.

Example 2

$3{x}^{2}+2x+1=0$

Here, $a=3$ , $b=2$ , and $c=1$ . Substituting, we get:

$x=-2±\frac{\sqrt{{2}^{2}-4\left(3\right)\left(1\right)}}{2\left(3\right)}$

Simplify.

$x=-2±\frac{\sqrt{-8}}{6}$

Because the discriminant is negative (-8), there are no real solutions to this quadratic equation.

Example 3

${x}^{2}-4x+2=0$

Here $a=1$ , $b=-4$ , and $c=2$ . Substituting, we get:

$x=-\left(-1\right)±\frac{\sqrt{{\left(-4\right)}^{2}-4\left(1\right)\left(2\right)}}{2\left(1\right)}$

$x=4±\frac{\sqrt{16-8}}{2}$

$x=4±\frac{\sqrt{8}}{2}$

Simplify.

$x=4±\frac{\sqrt{4×2}}{2}$

$x=4±2\frac{\sqrt{2}}{2}$

$x=\frac{2\left(2±\sqrt{2}\right)}{2}$

$x=2±\sqrt{2}$

Because the discriminant is positive, but not a perfect square, we have two real solutions:

$x=2+\sqrt{2}$ and $x=2-\sqrt{2}$

## Flashcards covering the Quadratic Formula

Algebra 1 Flashcards