# 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`

Answer

Squaring both sides gives:

`3x+4 = x^2`

`x^2 - 3x - 4 = 0`

`(x-4)(x+1)=0`

So our solutions are `x = 4` and `-1`.

CHECK:

Substituting `x = 4` in our original equation gives:

`"LHS" = sqrt(3(4) + 4)=sqrt16 = 4`

`"RHS" = 4`

Checks OK

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

`"LHS" = sqrt(3(-1) + 4)=sqrt1 = 1`

`"RHS" = -1`

DOES NOT check OK

So we conclude the only solution is `x = 4`.

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

Answer

Squaring both sides gives:

`13+sqrt(x)=x+2 sqrt(x) + 1`

Subtract `x+sqrt(x)+1` from both sides:

`12-x=sqrt(x)`

Squaring both sides again gives:

`144-24x+x^2=x`

Subtract `x` from both sides.

`x^2-25x+144=0`

Split `25` into `16` and `9`, which multiply to give `144`.

`x^2-16x-9x+144=0`

`x(x-16)-9(x-16)=0`

`(x-9)(x-6)=0`

This gives us `x = 9` or `x = 16`.

CHECK:

Substituting `x = 9` in our original equation gives:

`"LHS" `

`= sqrt(13+sqrt(9)) `

`= sqrt(13+3)`

`=sqrt(16)`

`=4="RHS" `

Checks OK

Substituting `x = 16` in our original equation gives:

`"LHS" `

`= sqrt(13+sqrt(16)) `

`= sqrt(13+4)`

`=sqrt(17)`

`"RHS" `

`=sqrt(16)+1`

`=4+1=5`

DOES NOT check OK

So we conclude the only solution is `x = 9.`

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

Answer

Squaring both sides gives:

`sqrt(x+10)=x-2`

Squaring both sides again gives:

`x+10=x^2-4x+4`

`x^2-5x-6=0`

`(x+1)(x-6)=0`

This gives us `x = -1` or `x = 6`.

CHECK:

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

`"LHS"=root(4)(-1+10)=root(4)(9)`

`"RHS"=sqrt(-1-2)=sqrt(-3)`

Does not check OK

Substituting `x = 6` in our original equation gives:

`"LHS"=root(4)(6+10)=root(4)(16)=2`

`"RHS"=sqrt(6-2)=sqrt(4)=2`

Checks OK

So we conclude the solution is `x = 6`.

Easy to understand math videos:

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**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.]

Answer

To make the writing much easier, let

`R_1^2=X` `R_2^2=Y` `r_1^2=x` `r_2^2=y`

So the equation becomes:

`kC=sqrt(X-Y)+sqrt(x-y)-A`, and we now need to solve for
*x*.

Add *A* to both sides; subtract `sqrt(X-Y)` from both sides:

`kC+A-sqrt(X-Y)=sqrt(x-y)`

and then square both sides:

`(kC+A-sqrt(X-Y))^2=x-y`

Re-arrange:

`x=y+(kC+A-sqrt(X-Y))^2`

We now re-express it in the original notation:

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

We have solved the equation for `r_1^2`, as required. There is no need to expand out the last bracket on the RHS.

Easy to understand math videos:

MathTutorDVD.com

**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.

Answer

We need to find `m` in `v = sqrt((2E)/m)` given that `v=25` and `E = 100,000`.

Substituting:

`25 = sqrt((200000)/m)`

Squaring both sides:

`625 = 200000/m`

Taking reciprocal of both sides:

`1/625 = m/200000`

Multiplying both sides by `200000`, we obtain

`m = 200000/625 = 320\ "kg"`.

Checking our answer in the original equation gives:

`v = sqrt((200000)/320) = 25`. Checks OK!

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