Question

How do you solve `sqrt(-10x - 4) = 2x`?

Answer 1

This has no answer. At first glance:

1. Square it (including every sign outside the square root).
2. Bring everything else to the other side with the 2nd degree term.

You get:
`4x^2 + 10x + 4 = 0`

Divide by 2.
`2x^2 + 5x + 2 = 0`

Factor.
`(2x+1)(x+2) = 0`

`x = -1/2, x = -2`

Check:

`sqrt(-10(-1/2)-4) = 2(-1/2) =>` 1 is not equal to -1. This answer is extraneous, even though -1 is equal to -1 by saying the square root yields the positive and negative answer. One of them is false, so the result is contingently true.

`sqrt(-10(-2)-4) = 2(-2) =>` 4 is not equal to -4. So is this one, even though -4 is equal to -4 by saying the square root yields the positive and negative answer. One of them is false, so the result is contingently true.

As-written, there are no solutions. Math at its core is pure logic that is supposed to always be true if done correctly, not just sometimes . There has to be a typo on the sign on the left, or it's a trick question. It should be:

`-sqrt(-10x-4) = 2x`

if there is to be an answer. This is a result of isolating one result of a square root. It has to be a `pmsqrt(stuff)` result, and only one of them had a real solution.

Answer 2

This equation has no solutions.

Though there could be a typo and it could be:

`-sqrt(-10x-4) = 2x`

We can square both sides to get

`-10x-4 = (2x)^2`

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

This gives us a quadratic equation:

`4x^2+10x+4 = 0`

Dividing both sides by 2, we get:

`(4x^2+10x+4)/2 = 0/2`

`2x^2+5x+2 = 0`

We use the Splitting the Middle Term technique to factorise the expression on the left

`2x^2+4x+x+2 = 0`

`2x*(x+2)+1*(x+2) = 0`

`(2x+1)*(x+2) = 0`

This tells us that

`2x + 1 = 0` or `x+2 = 0`

`color(green)(x = -1/2` or `color(green)( x = -2`

x can take either of these values and both will satisfy the equation `- sqrt(-10x - 4) = 2x`