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

I just need to know if i am right, the answers are in parentheses. 1. Solve for x: (x - 9)^2 = 1 (x=10) 2. Solve for x: x^2 + 24x + 90 = 0 (x=-12\pm3\sqrt{6}) 3. Solve for x: 2x^2 - 4x - 14 = 0 (x=1\pm2\sqrt{2}) 4. Create your own quadratic equation and demonstrate how it would be solved by graphing, factoring, the quadratic formula, and by completing the square. (My quadratic equation is x^2 - 6x = 0 and x = 0)

OpenStudy (anonymous):

1) Is correct

jimthompson5910 (jim_thompson5910):

for #1, 10 is just one of the two solutions

jimthompson5910 (jim_thompson5910):

there is another you're missing

OpenStudy (mathstudent55):

1 is incomplete

jimthompson5910 (jim_thompson5910):

#2 is correct, nice job

OpenStudy (anonymous):

1) could also be 8

jimthompson5910 (jim_thompson5910):

# 3 is also correct

OpenStudy (anonymous):

what about the last one?

jimthompson5910 (jim_thompson5910):

x^2 - 6x = 0 has 2 solutions as well x = 0 is one solution, but there is another

OpenStudy (anonymous):

what is the other answer? x = 8 ?

OpenStudy (anonymous):

Number 1 can be x=8

jimthompson5910 (jim_thompson5910):

x^2 - 6x = 0 has 2 solutions x = 0 or x = 6

jimthompson5910 (jim_thompson5910):

(x - 9)^2 = 1 has the two solutions: x = 10 or x = 8

OpenStudy (anonymous):

oh okay(: Thanks a lot! ^-^

jimthompson5910 (jim_thompson5910):

np

OpenStudy (anonymous):

for the first one this is what i did (x−9)2=1 ((x−9)2−−−−−−−√=1√ x-9=1 x=1+9,x=10 i got the answer since you guys said i got the right answer but did i do the work right?

OpenStudy (anonymous):

\[\sqrt{(x-9)^2} = \sqrt1\]

jimthompson5910 (jim_thompson5910):

(x-9)^2 = 1 x-9 = plus/minus sqrt(1) x-9 = sqrt(1) or x-9 = -sqrt(1) x-9 = 1 or x-9 = -1

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!