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Mathematics 21 Online
OpenStudy (anonymous):

How do you solve x(x-1)=4x

OpenStudy (amistre64):

one solution, if x not equal to 0, can be solved by dividing off an x

OpenStudy (amistre64):

if you know one solution, the other can be gotten by a variety of methods.

OpenStudy (akashdeepdeb):

Open the brackets/parenthesis like this. x(x-1)=4x x2 - x = 4x Now bring them down to 2 terms. You'll get a quadratic equation! Understood till here? :D

OpenStudy (anonymous):

Ok so you have x2-5x=0

OpenStudy (anonymous):

Then what you cannot factor

OpenStudy (akashdeepdeb):

Awesome. Now factorize this. What do you get?

OpenStudy (akashdeepdeb):

x(x-5) = 0 That IS factorization.

OpenStudy (anonymous):

Yes!!!!

OpenStudy (anonymous):

Thanks

OpenStudy (amistre64):

x(x-1) = 4x ; x=0 is an obvious solution (id hope) x-1 = 4 , provides the other solution

OpenStudy (anonymous):

Got it 0 and 5

OpenStudy (akashdeepdeb):

Thus as x2-5x is a quadratic equation. It'll have 2 roots. So x2-5x = 0 x can be 0 OR it can be x-5 = 0/x x = 5 So FINALLY. x = 0,5 Understood? :)

OpenStudy (anonymous):

Zero is not a possibility Thank you for your help

OpenStudy (akashdeepdeb):

:)

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