SOLUTION: Solve the equation sqrt (10 + x) = -1 - sqrt (11 - 5x)

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Question 172938This question is from textbook Introductory Algebra
: Solve the equation
sqrt (10 + x) = -1 - sqrt (11 - 5x)
This question is from textbook Introductory Algebra

Found 2 solutions by solver91311, Mathtut:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt+%2810+%2B+x%29+=+-1+-+sqrt+%2811+-+5x%29

Square both sides:

10%2Bx=1%2B2sqrt%2811-5x%29%2B11-5x

Collect like terms:

-2%2B6x=2sqrt%2811-5x%29

Square both sides again:

4-24x-36x%5E2=4%2811-5x%29

Collect like terms and put into standard form:

36x%5E2-24x-40=0

Solve the quadratic:

In fact, this horror actually factors %2836x-40%29%28x%2B1%29=0. So x=10%2F9 or x=-1

Well, that was tidy. But there is a big problem. Neither one of the roots calculated actually satisfies the original equation -- try it for yourself. This makes sense because how could the positive square root on the left be equal to a clearly negative number on the right? What happened is that the process of squaring the variable (twice in this case) introduced extraneous roots. There is no solution to this equation.

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt+%2810+%2B+x%29+=+-1+-+sqrt+%2811+-+5x%29

Square both sides:

10%2Bx=1%2B2sqrt%2811-5x%29%2B11-5x

Collect like terms:

-2%2B6x=2sqrt%2811-5x%29

Square both sides again:

4-24x-36x%5E2=4%2811-5x%29

Collect like terms and put into standard form:
:
solver was right up to this point
he zigged when he should have zagged...so easy to do in math...he forgot to add the 20x from the right side to the left side.
......so the final equation should be
:
36x%5E2-4x-40=0
:
9x%5E2-x-10=0reduced divided terms by 4
:
solutionssystem%28x=1.111111%2Cx=-1%29
:
both solutions work in original equation.
sqrt%2810-1%29=-1-sqrt%2811-5%28-1%29%29
3=3
:
sqrt%2810%2B1.111%29=-1-sqrt%2811-5%281.1111%29%29
3.33=3.33
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 9x%5E2%2B-1x%2B-10+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A9%2A-10=361.

Discriminant d=361 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+361+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+361+%29%29%2F2%5C9+=+1.11111111111111
x%5B2%5D+=+%28-%28-1%29-sqrt%28+361+%29%29%2F2%5C9+=+-1

Quadratic expression 9x%5E2%2B-1x%2B-10 can be factored:
9x%5E2%2B-1x%2B-10+=+%28x-1.11111111111111%29%2A%28x--1%29
Again, the answer is: 1.11111111111111, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+9%2Ax%5E2%2B-1%2Ax%2B-10+%29