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

2 Answers
Apr 10, 2015

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.

Apr 10, 2015

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#