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
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.