Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Solve the equation. square root of 5x+1=6 square root of x-2 +7=-4 x2+7x+4=0 5x2=-15

OpenStudy (anonymous):

1) x=7 2) x=123 3) x=-6.37, -.63 4) x=9

OpenStudy (anonymous):

how did u get these answers

OpenStudy (radar):

Would like to see how the answer to the 4th one was developed. when 9 is substitued in the equation: \[5(9)^{2}=-15\] \[(9)^{2}=-15/5=-3\] \[81\neq-3\]

OpenStudy (anonymous):

Yeah, my bad I read it wrong. It's actually 3i.

OpenStudy (radar):

Gettin there...... how about \[i \sqrt{3}\]

OpenStudy (anonymous):

\[5x ^{2}= -15\] \[X ^{2}= -15\div5=-3\] \[X=\sqrt{-3}\] \[X= i \sqrt{3}\]

OpenStudy (anonymous):

Sorry about that, I'm new to this and I hadn't yet figured out the equation inserts.

OpenStudy (radar):

What method did you use on number 3. Did you use quadratic formula?

OpenStudy (anonymous):

Yes. First I put it in the equation and solved, simplifying tona decimal and rounding to the nearest hundredth. I also checked the zeroes on the graph.

OpenStudy (anonymous):

5x^2 = -15 x^2 = -3 take the square root x = +- i sqrt(3)

OpenStudy (radar):

Very good. lerachia is there any more questions?

OpenStudy (radar):

blexting thanks , so their is two answers for number 4 \[i \sqrt{3}\]\[-i \sqrt{3}\]

OpenStudy (radar):

lerachia, do you understand how quantum torch got those answers or do you care?

OpenStudy (anonymous):

Yes, when you move an even power (either the numerator or denominator) there are multiple answers. I don't know how to write that with the equation thing yet though, but the rule applies, so thanks on my part as well.

OpenStudy (radar):

good night all.

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!