F.4 Maths. Equation

(a)4-4(1-k)<0
k<0
(b)k^2-4(2k-1)
=k^2-8k+4
k^2>0
-8k>0
so,k^2-8k+4>0
equation 2)has two unequal real root
(c)Result of (a),
k=-1
x^2+(-1)x = 1-2(-1)
x^2-x-3=0
x=(1+開方13)/2 or x==(1-開方13)/2

2) consider the equations:

1) x^2+2x = k-1
2) x^2+kx = 1-2k

where k is a constant.

(a) If the equation
1) has no real roots,find the range of values of k.
Ans:
Since has no real roots
Δ<0
2^2-4(1)(-k-1)<0
4+4k-4<0
k<0//

(b) Result of (a), show that equation 2)has two unequal real root.
Δ
=b^2-4ac
=k^2+4(1-2k)
=k^2+4-8k
=(k^2-4k+4)-4k
=(k-2)^2-4k >0

(c) Solve equation 2) if k is the largest integer in obtained in (a)
the largest integer in obtained in (a) is -1

So x^2+kx = 1-2k
=> x^2-x=3
=>x^2-x-3=0
=>x=2.84 or 3.27//