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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
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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
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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? :)