4 Quadratics and polynomials

Lesson

We call equations like $x-7=2$`x`−7=2 linear equations. These are equations where all variables have a power of $1$1. Linear equations have only one solution. That is, there is only one value of the variable which will make the equation true. In this case, the only solution is $x=9$`x`=9

If instead, the variables in the equation have a power of $2$2, we call them quadratic equations. Quadratic equations can potentially have two solutions. For example, in the equation $x^2-7=2$`x`2−7=2, there are two solutions, $x=3$`x`=3 and $x=-3$`x`=−3.

Solve $x^2+4=20$`x`2+4=20 for $x$`x`.

Following our rules for solving linear equations, we want to isolate $x$`x` and whatever we do to one side of the equation we do to the other. So our first step is to subtract $4$4 from both sides of the equation, giving us $x^2=16$`x`2=16.

Next we want to undo raising $x$`x` to the power of $2$2. However, we need to be careful here. There are two operations which could be the reverse of squaring a number. We have to take both the positive and negative square root. This will give us two solutions.

$x^2$x2 |
$=$= | $16$16 | |

$x$x |
$=$= | $\pm\sqrt{16}$±√16 |
Taking the positive and negative square root of both sides |

$=$= | $\pm4$±4 |
Evaluating the positive and negative square roots |

The symbol $\pm$± means "plus or minus". We can use this as a shorthand for both the positive and negative of a number. In this case, it means that our solutions are $x=4$`x`=4 and $x=-4$`x`=−4.

We can check these solutions by substituting them in to the original equation and seeing if it holds true.

The null factor law states that if a product of two or more factors is equal to $0$0, then at least one of those factors must be equal to $0$0. For example, if $xy=0$`x``y`=0 then either $x=0$`x`=0 or $y=0$`y`=0. We can use this to solve quadratic equations.

Solve $5\left(x-2\right)\left(x+3\right)=0$5(`x`−2)(`x`+3)=0 for $x$`x`.

We have a product of three factors equal to $0$0, so by the null factor law one of these factors must be equal to $0$0. Since $5$5 is a constant, we know that $5\ne0$5≠0. Only factors that contain a variable can be equal to $0$0 (unless the constant factor is $0$0 itself).

This leaves two possibilities. Either $x-2=0$`x`−2=0 or $x+3=0$`x`+3=0. We've converted our quadratic equation into two linear equations. Solving each linear equation for $x$`x` gives us $x=2$`x`=2 and $x=-3$`x`=−3, so these are the solutions to the original equation.

Once again, we can check these solutions by substituting them in to the original equation and seeing if it holds true.

Did you know?

We call raising a number to the power of $2$2 "squaring" the number. "Quadratic" comes from the ancient Latin for "square". So quadratic equations can be thought of as square equations.

Summary

Equations where all of the variables have a power of one are linear equations. These have at most one solution.

Equations where some of the variables have a power of two are quadratic equations. These have at most two solutions.

The symbol $\pm$± means "plus or minus".

If we can rearrange a quadratic equation into the form $x^2=k$`x`2=`k`, then we can solve the equation by taking the positive and negative square roots. That is, $x=\pm\sqrt{k}$`x`=±√`k`.

The null factor law states that if a product of two or more factors is equal to $0$0, then at least one of those factors must be equal to $0$0.

If we can rearrange a quadratic equation into the form $\left(x-a\right)\left(x-b\right)=0$(`x`−`a`)(`x`−`b`)=0 then we know that either $x-a=0$`x`−`a`=0 or $x-b=0$`x`−`b`=0. We can solve the quadratic equation by solving each of these linear equations.

Solve $x^2=2$`x`2=2 for $x$`x`.

Enter each solution as a surd on the same line, separated by a comma.

Solve $\frac{x^2}{16}-2=2$`x`216−2=2 for $x$`x`.

Enter each solution on the same line, separated by a comma.

Solve $\left(x-6\right)\left(x+7\right)=0$(`x`−6)(`x`+7)=0 for $x$`x`.

Enter each solution on the same line, separated by a comma.

Solve simple quadratic equations using a range of strategies.