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!
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 .