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Mathematics 21 Online
OpenStudy (anonymous):

(x^2 - x -2)^3 how to solve this equation

Parth (parthkohli):

Equation?

OpenStudy (unklerhaukus):

an equation has an equal sign

OpenStudy (anonymous):

ok y=(x^2-x-2)^3

Parth (parthkohli):

lol do you have to solve for \(\rm x\)?

OpenStudy (anonymous):

i need the roots

OpenStudy (unklerhaukus):

factorise what is in the brackets

Parth (parthkohli):

To find the roots, you have to find the value of x where the function is 0.\[\Huge \rm (x^2 - x - 2)^3 = 0\]

OpenStudy (anonymous):

what next

Parth (parthkohli):

Find the cube root of both sides. Solve the quadratic equation.

OpenStudy (anonymous):

k thnx

OpenStudy (anonymous):

(x^2-x-2)^3 (x^2-2x+x-2)^3 (x(x-2)+1(x-2))^3 (x+1)^3 .(x-2)^3

Parth (parthkohli):

So what do you get?

OpenStudy (anonymous):

same as nitz

OpenStudy (anonymous):

thnx nitz

Parth (parthkohli):

That's not what I told you. lol

OpenStudy (anonymous):

i was jst unsure whether to open the brackets or not !!

Parth (parthkohli):

Well, it's OK. Her method works too!

OpenStudy (anonymous):

hahaha..yea it does

OpenStudy (anonymous):

thnx again

OpenStudy (unklerhaukus):

\[y=(x^2-x-2)^3\]\[0=(x^2-x-2)^3\]\[0=\left((x+1)(x-2)\right)^3\]\[0=(x+1)(x-2)\] \[x=-1,2\]

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