how would u do this problem 2+square root 19-2x

Question

zepdrix · Answer

$$\large 2+\sqrt{19-2x}=?$$Is this what the problem looks like? Solve for x? Or just simplify?

What is on the other side of the equation? :O

anonymous · Answer

yes that is the problem

anonymous · Answer

solve for x

zepdrix · Answer

=0?

anonymous · Answer

your just bsolving for x

zepdrix · Answer

you can't solve for x if we don't have an equality sign somewhere... :o

anonymous · Answer

it says x= that...would the problem just be undefined?

zepdrix · Answer

$$\large 2+\sqrt{19-2x}=x$$Ok that's what we needed to know.... So this is what it should look like?

anonymous · Answer

yes

zepdrix · Answer

Hmm this one is a bit tricky :) We'll have to do a bunch of algebra to solve for x.
First, subtracting 2 from each side gives us,$$\large \sqrt{19-2x}=x-2$$Then we'll square both sides to get rid of that square root,$$\large 19-2x=(x-2)^2$$

zepdrix · Answer

Then you need to expand out that binomial :D Remember how to do those?$$\large (x-2)^2\qquad =\qquad (x-2)(x-2)\qquad =\qquad x^2-2x-2x+4$$

anonymous · Answer

so the final answer is x=-3 and x=5 correct?

anonymous · Answer

(-3,0) and (5,0)

anonymous · Answer

?

zepdrix · Answer

Ah yes very good! :)

Sorry I had to go earlier D':