Quadratic Equation

1. 4/(2-x) = 6 - 5/(x+1)
4(x+1) = 6(2-x)(x+1) - 5(2-x)
4x+4 = 12x - 6x^2 + 12 - 6x - 10 + 5x
4x+4 = 11x - 6x^2 + 2
6x^2 - 7x + 2 = 0
(3x - 2)(2x - 1) = 0
x = 2/3 or x = 1/2
2) 4x^2 - 4x + p-1 = 0 has real roots,
△ = 4^2 - 4(4)(p-1) >= 0
16 - 16p + 16 >= 0
1 - p + 1 >= 0
p =< 2
the maximum value of p = 2
3a)Since it touches the x-axis (equation is y = 0) at only one point ,
y = 0..........(1)
y=x^2+(k+1)x+2k+7............(2)
(1) and (2) have 1 set solution only ,
sub (1) to (2):
x^2 + (k+1)x + 2k+7 = 0 have double root, so
△ = (k+1)^2 - 4(2k+7) = 0
k^2 + 2k + 1 - 8k - 28 = 0
k^2 - 6k - 27 = 0
(k + 3)(k - 9) = 0
k = - 3 or k = 9
b)When k = -3 :
x^2 + (-3+1)x + 2(-3) + 7 = 0
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1(double root)
When k = 9
x^2 + (9+1)x + 2*9 + 7 = 0
x^2 + 10x + 25 = 0
(x + 5)(x + 5) = 0
x = - 5 (double root)

Answer:

1. 4/2-x=6-5/x+1
2x-x^2=6x-5+x
x^2+5x-5=0
By using quadratic formula,
x=-5.854... or x=0.854...

2.△&gt;0
(-4)^2-4(4)(p-1)&gt;/0
16-16p+16&gt;/0
16p&lt;/32
p&lt;/2
So the maximum value of p is 2

3.A)△=0
(k+1)^2-4(1)(2k+7)=0
k^2+2k+1-8k-28=0
k^2-6k-27=0
(k+3)(k-9)=0
k=-3 or k=9

B)when k=-3,
x^2+(-3+1)x+2(-3)+7=0
(x-1)^2=0

So, x=1
when k=9,
x^2+(9+1)x+2(9)+7=0
(x+5)^2=0

So, x=-5