maths quadratic

1.
2^(2x+2)=9(2^x)-2
4(2^x)^2-9(2^x)+2=0
(2^x-2)(4*2^x-1)=0
2^x=2 OR 2^x=1/4
x=1 OR -2

2.
[root(x+9)]+11=x
x+9=(x-11)^2
x+9=x^2-22x+121
x^2-23x+112=0
(x-16)(x-7)=0
x=16 OR 7 (rejected)
therefore x=16

3.
Let s be the common root.
Then as^2+bs+c=0............(1)
ps^2+qs+r=0....................(2)

(1)*p-(2)*a,

pas^2+pbs+pc-(pas^2+qas+ra) = 0
s(pb-qa) + pc-ra=0
s=(cp-ar) / (aq-bp)..................(3)

(1)*q - (2)*b,

qas^2+qbs+qc-(pbs^2+qbs+rb)=0
s^2(qa-pb) + (qc - rb) = 0
s^2 = (br-cq) / (aq-bp)...............(4)

Sub (3) into (4),
[(cp-ar) / (aq-bp)]^2=(br-cq) / (aq-bp)
(cp-ar)^2 =(br-cq)*(aq-bp)

1.
Let 2^x = y.
So 2^(2x + 2) = (2^2x)(2^2) = 4y^2. So the equation becomes
4y^2 = 9y - 2
4y^2 - 9y + 2 = 0
(4y-1)(y-2) = 0
y= 1/4 or 2. That is 2^x = 1/4 or 2^x= 2.
2^x = 1/4 = (1/2)^2 = [2^(-1)]^2 = 2^(-2), so x = -2.
2^x = 2, so x = 1.
2.
sqrt(x+9) + 11 = x
sqrt(x+9) = x - 11
x + 9 = (x-11)^2 = x^2 - 22x + 121
x^2 - 23x +112 = 0
(x-7)(x-16) = 0
x= 7 (reject because it does not fit the equation) and x = 16.
3.
Let m be the common root. Therefore,
am^2 + bm + c = 0................(1) And
pm^2 + qm + r = 0.................(2)
(1) x p - (2) x a, we get
bpm + cp - aqm - ra = 0
(bp - aq)m = ra - cp
m = (ra-cp)/(bp - aq)...............(3)
(1) x q - (2) x b, we get
aqm^2 - bpm^2 + cq - br = 0
m^2(aq - bp) = (br - cq)
m^2 = (br-cq)/(aq - bp) ...........(4)
From (3) and (4)
(br-cq)/(aq-bp) = (ra - cp)^2/(bp - aq)^2
So (br-cq)(aq- bp) = (ra- cp)^2.