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`.

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.`

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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`.

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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
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.

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!