Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (anonymous):

solve square(x+5) = square(x^2-15)

OpenStudy (anonymous):

Does "square(...)" mean square root of "..."?

OpenStudy (anonymous):

yes sorry @SithsAndGiggles

OpenStudy (anonymous):

\[\begin{align*} \sqrt{x+5}&=\sqrt{x^2-15}\\ x+5&=x^2-15\\ x^2-x-20&=0 \end{align*}\]

OpenStudy (anonymous):

Left side can be factored

OpenStudy (anonymous):

(x-5)(x+4) @SithsAndGiggles

OpenStudy (anonymous):

Yep. \[(x-5)(x+4)=0~\Rightarrow~x=\cdots\]

OpenStudy (anonymous):

4? @SithsAndGiggles

OpenStudy (anonymous):

since it cant be negative right

OpenStudy (anonymous):

For the left equation to be true, you must have either \[x-5=0~~~\text{or}~~~x+4=0\] Or equivalently, \(x=5\) or \(x=-4\). The 4 can be negative because when you plug it into the original equation, you have \[\sqrt{-4+5}=\sqrt{(-4)^2-15}\\ \sqrt{1}=\sqrt{16-15}\\ 1=1\]

OpenStudy (anonymous):

thank you! @SithsAndGiggles

OpenStudy (anonymous):

you're welcome!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!