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Mathematics 7 Online
OpenStudy (anonymous):

Determine if the complex number, 2+ i, is a solution to the quadratic equation 3x2 - 6x + 6 = 0 Explain your answer.

OpenStudy (misty1212):

hi easiest just to solve if and see what you get

OpenStudy (misty1212):

\[3x^2-6x+6=0\] divide by 3 and solve \[x^2-2x+2=0\] you know how to solve this one?

OpenStudy (anonymous):

Plug in 2 + i to your equation

OpenStudy (misty1212):

probably easiest in this case to complete the square because 2 is even it will be real annoying to try to evaluate it at \(x=2+i\)

OpenStudy (misty1212):

the answer will probably be no in any case

OpenStudy (anonymous):

Can you please explain why it is not a solution

OpenStudy (misty1212):

sure when we solve it we will get something else!

OpenStudy (anonymous):

Okay thank you . Could you help me with one more?

OpenStudy (misty1212):

\[x^2-2x+2=0\\ x^2-2x=-2\\ (x-1)^2=-2+1=-1\\ x-1=\pm i\\ x=1\pm i\]

OpenStudy (misty1212):

the solutions are \(1\pm i\) not \(2\pm i\)

OpenStudy (misty1212):

sure why not?

OpenStudy (anonymous):

Determine if 1 - i is a solution to the equation 2x2 - 4x + 4 = 0. Explain your answer.

OpenStudy (misty1212):

lol this is really sooo much different than the last one why don't they just ask you to solve it instead of screwing around with "is blah blah..."

OpenStudy (misty1212):

\[2x^2-4x+4=0\] divide by 2 \[x^2-2x+2=0\] subtract 2 \[x^2-2x=-2\] complete the square \[(x-1)^2=-2+1=-1\] take the square root \[x-1=\pm i\] solve for \(x\)\[x=1\pm i\]

OpenStudy (misty1212):

so this time the answer is YES

OpenStudy (anonymous):

Thank you so much!

OpenStudy (misty1212):

you are welcome dear glad to help!

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