# 6. Equations With Radicals

It is important in this section to **check your solutions** in the original equation, as the process that we use to solve these
often produces solutions which actually don't work when subsituted back into the original equation.

(In fact, it is **always** good to check solutions for equations - you learn so much more about why things work the way they do. ^_^)

## Example

Solve for `x`: `sqrt(2x+6)=2x`

Answer

Squaring both sides gives:

`2x + 6 = 4x^2`

`4x^2 - 2x - 6 = 0`

`2x^2 - x - 3 = 0`

`(2x-3)(x+1)=0`

So our solutions are `x = 1.5` or `-1`.

CHECK:

Substituting `x = 1.5` in our original equation gives:

`"LHS" = sqrt(2(1.5) + 6) = sqrt9 = 3`

`"RHS" = 2 times 1.5 = 3`

Checks OK

Substituting `x = -1` in our original equation gives:

`"LHS" = sqrt(2(-1) + 6)=sqrt4 = 2`

`"RHS" = 2 times -1 = -2`

DOES NOT check OK

So we conclude the only solution is `x = 1.5`.

Can you figure out **where** the 'wrong' solution is coming from?

### Exercises

**1. **Solve** **`sqrt(3x+4)=x`

**2.** Solve `sqrt(13+sqrtx)=sqrtx+1`

**3. **Solve `root(4)(x+10)=sqrt(x-2)`

**4. **(Application) In the study of spur gears in contact, the following
equation arises:

`kC=sqrt(R_1^2-R_2^2)+sqrt(r_1^2-r_2^2)-A`

Solve for `r_1^2`.

Spur gears [image source]

[Being able to solve equations like this is important for science and engineering. We often need to solve for a particular variable in a formula.]

**5. **(Application) The velocity of an object with a known mass and known Kinetic Energy is given by:

`v = sqrt((2E)/m)`, where

`v` is the velocity in `"ms"^-1`

`E` is the kinetic energy, measured in joules

`m` is the mass in kg

Find the mass of an object that is moving at `25\ "ms"^-1`, given that the kinetic energy is `100,000` joules.

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