Show that x = 2 is on root of the equation x^e -4x^2 + (q + 4)x - 2q = 0.

If the other two roots of this equation differ by 4, find the value of q, and solve the equation completely.

Can someone show me how to do this q please? Thanks!

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- Dec 15th 2009, 03:14 PMDetanonCubic Equation
Show that x = 2 is on root of the equation x^e -4x^2 + (q + 4)x - 2q = 0.

If the other two roots of this equation differ by 4, find the value of q, and solve the equation completely.

Can someone show me how to do this q please? Thanks! - Dec 15th 2009, 04:16 PMmathaddict
HI

do u mean x^3 instead of x^e for the leading term ? Because if its x^e , you will end up with $\displaystyle 2^e-2^3\neq$ 0 .

Assuming that its $\displaystyle x^3-4x^2+(q+4)x-2q=0$

$\displaystyle (x-2)(x^2-2x+q)=0$

Let**a**and**b**be the 2 other roots .

a+b=2

ab=q

a-b=4

So now there are enough information for you to solve for q .