maths. F.4

1: slove the equation (x-3)2 -(2x-5)2=0
(x-3)2 =(2x-5)2

x-3=2x-5 or x-3=-(2x-5)
x=2 or x=8/3

2: suppose -k is a root of the equation x^2 -3x+(k+12)=0. find the
possible values of -k.

(-k)2 -3(-k)+(k+12)=0
k2+4k+12=0
(k+6)(k-2)=0
k=-6 or k=2
-k=6 or -k=-2


3: the equation 2x2 -4(x-1)+k=0 has 2 distinct real roots, find the r
ange of k.
△>0
[-4(x-1)]2-4(2)(k)>0
16(x-1)2-8k>0
16(x2-2x+1)-8k>0
2(x2-2x+1)-k>0
2x2-5x+2>0
x2-5/2x+1>0
x2-5/2x+25/16>9/16
(x-5/4)2>9/16
x-5/4<-3/4 or x-5/4>3/4
x<1/2 or x=2


4: slove the quadratic equation (1-x)(4+x)+9x-2=0
(1-x)(4+x)+9x-2=0
4+x-4x-x2+9x-2=0
2+6x-x2=0
x2-6x-2=0
x2-6x+9=11
(x-3)2=11

x-3=√11 or x-3=-√11
x=3+√11 or x=3-√11

5: if the equation (x+1)(x-4)+kx-4k=0 has a repeated root, find the
possible values of k.
(x+1)(x-4)+kx-4k=0
x2-3x-4+kx-4k=0

i.e. x2+(k-3)x-(4k+4)=0

△=0

(k-3)2-4(1)[-(4k+4)]=0
k2-6k+9+16k+16=0

k2+10k+25=0
(k+5)2=0
k=-5( repeated)


7: suppose f(x) =x^2-4x. If f(k-2)=f(k), find the value of k.
f(k-2)=f(k)
(k-2)2-4(k-2)=k2-4k

k2-4k+4-4k+8=k2-4k
-4k+12=0
k=3



9: consider the graph of y= x^2-8x+25
(a) find the vertex of the graph
(b) find the axis of symmetry

(a)
y= x2-8x+25
=x2-8x+16+9
=(x-4)2+9
the vertex of the graph=(4,9)
(b)
x=4


11: If log 2.5=k, express each of the following in terms of k.
(a) log 25
(b) log 0.4

(a) log 25
=log(2.5 x 10)
=log2.5+log10
=k+1