Question

Given that p and 2 are the roots of the quadratic equation q(1-x)=(x+3)(1-2x),where q is a constant,calculate the values of p and of q.
@rational @radar

Answer 1

\[q-qx=2x^2-x+3-6x\]\[-2x^2+7x-qx+q-3=0\]

Answer 2

\[(x+3)(1-2x)=x-2x^2+3-6x\]

Answer 3

ops,thnx @mustafa2014

Answer 4

np :)

Answer 5

\[2x^2+5x-qx+q-3=0\]

Answer 6

divide both side by 2

Answer 7

q-qx=x-2x^2+3-6x
q-qx-x+2x^2-3+6x=0
x^2+(5-q)/2x+(q-3)/2=0
-2-p=(5-q)/2
-4-2p=5-q
q-2p=9
2p=(q-3)/2
4p-q=-3
4p-9-2p=-3
2p=6
p=3
q=15
x^2-5x+6=0

Answer 8

i don't get it. why -2-p?

Answer 9

cause x^2+(5-q)/2x+(q-3)/2=0
-2-p=(5-q)/2
-4-2p=5-q
q-2p=9
2p=(q-3)/2

Answer 10

p and 2 are the roots means (x-p)(x-2)=0
x^2-2x-px+2p=0

Answer 11

x^2-(2+p)x+2p=0
and we have
x^2+(5-q)/2x+(q-3)/2=0

Answer 12

Thnx for ur explanation @mustafa2014 :)

Answer 13

np :)