**Solving Nonlinear Equations** *KS3 Revision*

## What you need to know

**Things to remember:**

- We can find the value(s) that satisfy an equation by rearranging to get the variable by itself.

We have seen linear equations before, they’re equations that can be drawn as a straight line (they don’t have powers). So, they usually look something like this:

y=5x+2

y=7-3x

8y+5x=3

A nonlinear equation, however, cannot be drawn as a straight line. These equations involve powers and multiplying variables together. They look something like this:

y=x^2+4

y=\sqrt{x}

xy=8

We’re going to look at two types of questions here, and each will take two steps.

*Solve the following nonlinear equation x^2+4=29.*

** **

**Step 1:** Get the variable by itself.

*Hint:** Remember, if we see an addition we subtract it, and if we see plus we subtract it.*

* *

x^2+4=29

x^2+4-4=29-4

x^2=25

* *

**Step 2: **Do the opposite of what you’re doing to the x.

*Hint: **What is the opposite of square? Square rooting!*

* *

x^2=25

\sqrt{x^2}=\sqrt{25}

x=5

*Note: **A square root provides two answers, a positive and a negative (-5\times-5=25.*

* *

*Solve the following nonlinear equation \sqrt{x}-3=9.*

** **

**Step 1:** Get the variable by itself.

*Hint:** Remember, if we see an addition we subtract it, and if we see plus we subtract it.*

* *

\sqrt{x}-3=9

\sqrt{x}+3=9+3

\sqrt{x}=12

* *

**Step 2: **Do the opposite of what you’re doing to the x.

*Hint: **What is the opposite of square rooting? Squaring!*

* *

\sqrt{x}=12

\sqrt{x}^2=12^2

x=144

## Example Questions

**Question 1:** *Solve the following nonlinear equation x^2-3=13.*

**Step 1:** Get the variable by itself.

*Hint:** Remember, if we see an addition we subtract it, and if we see plus we subtract it.*

* *

x^2-3=13

x^2+3=13 +3

x^2=16

* *

**Step 2: **Do the opposite of what you’re doing to the x.

*Hint: **What is the opposite of square? Square rooting!*

* *

x^2=16

\sqrt{x^2}=\sqrt{16}

x=4

* *

*Note: **A square root provides two answers, a positive and a negative (-4\times-4=16.)*

**Question 2:** *Solve the following nonlinear equation \sqrt{x}+8=28.*

**Step 1:** Get the variable by itself.

*Hint:** Remember, if we see an addition we subtract it, and if we see plus we subtract it.*

* *

\sqrt{x}+8=28

\sqrt{x}+8-8=28-8

\sqrt{x}=20

* *

**Step 2: **Do the opposite of what you’re doing to the x.

*Hint: **What is the opposite of square rooting? Squaring!*

* *

\sqrt{x}=20

\sqrt{x}^2=20^2

x=400