# 10.03 Solving quadratic equations using square roots

Lesson

Solving basic or simple quadratics involves using algebraic manipulation. Let's look at a few examples of this process.

Remember!

We will be manipulating expressions with squared terms. Remember that when we take the square root of both sides of an equation we need to consider both the positive and negative square roots.

 $x^2$x2 $=$= $25$25 $x$x $=$= $\pm\sqrt{25}$±√25 $x$x $=$= $\pm5$±5

Both $x=5$x=5 and $x=-5$x=5 are solutions because when squared they both give us $25$25.

#### Worked examples

##### Question 1

Solve the quadratic equation $x^2-9=0$x29=0.

Think: We want to end up with an expression like $x=\dots$x=, and to get there we will need to isolate $x$x

Do: Considering the order of operations, we will first move the constant term to the other side, then undo the square:

 $x^2-9$x2−9 $=$= $0$0 $x^2$x2 $=$= $9$9 $x$x $=$= $\pm\sqrt{9}$±√9 $x$x $=$= $\pm3$±3

Since there are two numbers that, when squared, give $9$9, the solutions to this equation are $x=3$x=3 or $x=-3$x=3.

##### Question 2

Find solutions to $\left(2x+7\right)^2=64$(2x+7)2=64.

Think: Identify the order of operations necessary to isolate the $x$x variable. In this case we will deal with the square root first, then the addition and finally the multiplication by $2$2.

Do:

 $\left(2x+7\right)^2$(2x+7)2 $=$= $64$64 $2x+7$2x+7 $=$= $\pm\sqrt{64}$±√64

We use inverse operations to remove the square from the $LHS$LHS, the opposite of a square is a square root. Next we have

 $2x+7$2x+7 $=$= $\pm8$±8 (from here on, this is just like solving linear equations)

Where we have evaluate the square root. There are two solutions now, let's continue to solve both for $x$x:

 $2x+7$2x+7 $=$= $8$8 or $2x+7$2x+7 $=$= $-8$−8 $2x$2x $=$= $8-7$8−7 or $2x$2x $=$= $-8-7$−8−7 $2x$2x $=$= $1$1 or $2x$2x $=$= $-15$−15 $x$x $=$= $\frac{1}{2}$12​ or $x$x $=$= $\frac{-15}{2}$−152​

##### Question 3

Find the solutions to $x^2-17=0$x217=0.

Think: Identify the order of operations necessary to isolate the $x$x variable. In this case we will deal with the subtraction first, then the square.

Do:

 $x^2-17$x2−17 $=$= $0$0 $x^2$x2 $=$= $17$17

Here we use inverse operations to remove the $-17$17 from the $LHS$LHS, the opposite of a $-17$17 is a $+$+ $17$17. Then we have

 $x$x $=$= $\sqrt{17}$√17 OR $x$x $=$= $-\sqrt{17}$−√17

Where we use inverse operations to remove the square from the $LHS$LHS, the opposite of a square is a square root.

### A special note about exact solutions

For nearly all of our work with solutions to functions and equations it is standard practice to leave our final expression in exact form. In this case, $\sqrt{17}$17 is as far as we will go, we cannot simplify the root any further.

In questions involving applications of quadratics, we may be asked to evaluate the square root at the very end and then approximate to a specific number of decimal places. For example, perhaps we want the time taken to travel a certain distance or an estimate of the area of a block of land.

### A special note about number of solutions

All of the worked examples above had two solutions, but let's consider some other possibilities.

Let's consider the simplest quadratic to solve $x^2=a$x2=a. We will have three cases for $a$a

• $a>0$a>0, that is a is positive, for example $x^2=16$x2=16
• $a=0$a=0, for example $x^2=0$x2=0
• $a<0$a<0, that is a is negative, for example $x^2=-16$x2=16

Consider how many solutions there are to each of the scenarios above. What happens when you take the square root of both sides?

• $x^2=a$x2=a, $a>0$a>0, for example $x^2=16$x2=16, will give $2$2 solutions $x=\sqrt{a}$x=a and $x=-\sqrt{a}$x=a
• $x^2=0$x2=0, for example $x^2=0$x2=0, will have $1$1 solutions $x=0$x=0
• $x^2=a$x2=a, $a<0$a<0, for example $x^2=-16$x2=16, will have no real solutions as we cannot square root a negative in the real numbers
Remember!

A quadratic equation can have $0$0, $1$1 or $2$2 solutions. We will explore this further with graphing quadratics.

#### Practice questions

##### Question 5

Solve for $p$p:

$5\left(p^2-3\right)=705$5(p23)=705

##### Question 6

Solve $\left(x-4\right)^2=10$(x4)2=10 for $x$x.

1. Write all solutions on the same line, separated by commas.

### Outcomes

#### A1:A-REI.B.4

Solve quadratic equations in one variable.

#### A1:A-REI.B.4b

Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as "no real solution."