Question

Solve the equation. If there are two actual solutions (check) write them in set brackets, such as {-1,2} with no spaces. http://ncvps.blackboard.com/courses/1/Algebra1Sec05Fall11/ppg/respondus/exam_Quadratics_HW9/3329.gif

Answer 1

I don't think that page is publicly accessible?

Answer 2

\[\sqrt{2x+3=}\]

Answer 3

\[\sqrt{2x+3=x}\]

Answer 4

? :)

Answer 5

Haha sorry mate, distracted.

\[\Large \begin{array}{l}
\sqrt {2x + 3} = x\\
2x + 3 = {x^2}\\
2x = {x^2} - 3\\
- {x^2} + 2x + 3 = 0\\
- (x - 3)(x + 1) = 0
\end{array}\]
So we have the roots of 3 and -1, meaning x could be either. However, there is the constraint:\[\Large \begin{array}{l}
2x + 3 \ge 0\\
2x \ge - 3\\
x \ge - \frac{3}{2}
\end{array}\]Therefore we can disregard -1 since it is less than -(3/2), meaning x = 3. Understand?

Answer 6

Nope . :(