Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

What are the approximate solutions of 2x2 - 4x = 1 rounded to the nearest hundredth?

OpenStudy (campbell_st):

2x^2 - 4x = 1 or x^2 - 2x = 1/2 complete the square x^2 - 2x + 1 = 1/2 + 1 (x - 1)^2 = 3/2 \[x - 1 = \sqrt{3/2}\] \[x = 1\pm \sqrt{3/2}\] now evaluate the 2 solutions

OpenStudy (anonymous):

no real soulitons?

OpenStudy (campbell_st):

yep there are real solutions.... but you need a calculator to get the answer \[x = 1 + \sqrt{3/2}\] and \[x = 1-\sqrt{3/2}\]

OpenStudy (anonymous):

\[\LARGE \Delta =b^2-4ac\] if... \[\LARGE \Delta >0 \quad \text{two solutions.}\] \[\LARGE \Delta =0 \quad \text{one solution.}\] \[\LARGE \Delta <0 \quad \text{no solution }\]

OpenStudy (anonymous):

x ≈ -0.23 and x ≈ 2.23 ? right

OpenStudy (campbell_st):

yep... that makes sense

OpenStudy (anonymous):

how did you break it down

OpenStudy (anonymous):

No real solutions Two identical rational solutions Two different rational solutions Two irrational solutions What are the exact solutions of x2 − 5x − 1 = 0? Solve 2x2 + 3x + 6 = 0. Round solutions to the nearest hundredth. x ≈ −2.64 and x ≈ 1.14 x ≈ −1.14 and x ≈ 2.64 No real solutions x ≈ −1.11 and x ≈ −4.89 What are the approximate solutions of 2x2 − 4x = 1 rounded to the nearest hundredth? x ≈ −0.23 and x ≈ 2.23 x ≈ 1.71 and x ≈ 0.29 x ≈ −0.45 and x ≈ 4.45 No real solutions Using graphing technology, approximate the solutions of 0.55x2 − 0.33x − 5.6 = 0. x ≈ −2.91 and x ≈ 3.51 x ≈ −3.14 and x ≈ 3.81 x ≈ −2.76 and x ≈ 3.62 x ≈ −2.61 and x ≈ 3.84 Using graphing technology, approximate where the graph of y = 2x2 − 8x − 5 crosses the x-axis. (4.55, 0) and (−0.55, 0) (4.23, 0) and (−0.46, 0) (4.32, 0) and (−0.49, 0) (4.48, 0) and (−0.51, 0)

OpenStudy (anonymous):

DOES MY THING MAKE ANY SENSE AT 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!