EX 3.1
2.Determine the nature of the roots of
kx2 – x + k = 1 –2kx, where k  0 , and find the roots
in terms of k, where necessary
Solutions:
kx2 +2kx – x + k – 1 = 0
kx2 + x( 2k –1 ) - 1 = 0
a=k
b = 2k – 1
c=-1
D = b2 – 4ac
= ( 2k – 1 ) 2 – 4 (k) (k- 1)
= 4k2 – 4k +1 - 4 k2+4k
= 1 which is > 0
Solutions are
x
b D
2a
3.Find the value(s) or range of values of p for which the equation
a) px2 – 6x +p = 0 has equal roots,
a)Since the equation has real and equal roots , the discriminant , D = 0
a=p
b= –6
c=p
D = b2 – 4ac
=( - 6 ) 2 – 4 (p) (p)
= 36 – 4p2
D= 0
i.e 36 – 4p2
36
p2
p
=0
= 4p2
=9
=3
3b) 2x2 – 4x +3 = p has real roots
Since the equation has real roots , the discriminant D  0
Rewrite the given equation as follows
2x2 – 4x +3-p =0
a=2
b= –4
c = 3- p
D = b2 – 4ac 0
i.e (-4)2 – 4 (2 ) (3-p)  0
16 – 24 + 8p 0
-8 +8p 0
8p  8
p 1
11.If the roots of the equation px2 +qx +r = 0 are real , show that the roots of the
equation r2x2 + (2pr – q 2 )x +p2 = 0 are also real.
Since the equation
px2 +qx +r = 0 has real roots , the discriminant D  0
a=p
b= q
c=r
D = b2 – 4ac 0
i.e (q)2 – 4 (p ) (r)  0
q2 – 4p r  0 -----------------------(i)
Consider the equation
r2x2 + (2pr – q 2 )x +p2 = 0
a = r2
b = 2pr – q 2
c = p2
= b2 – 4ac
= (2pr – q 2)2 – 4 (r2)(p2)
= 4 r2p2 –4pq2r +q4– 4 r2p2
= –4pq2r +q4
= q2 (q2 – 4p r)  0
( because q2  0 for any value of q and q2 – 4p r  0 )
So, the equation px2 +qx +r = 0 also has real roots.
D
13 From the simultaneous equations 2x – 3y = 4 , 2x2 – 9y2 =k.
Derive an equation relating x and k. Hence find the range of values of k if the simultaneous equations
have real solution(s).
Solution:
2x – 3y = 4 -------------(1)
2x2 – 9y2 =k.------------(2)
From Eqn. (1) 3y = 2x –4
Substitute in eqn (2), we get
2x2 – ( 2x-4)2
2x2 – (4x2 –16x +16 )
– 2x2 +16x -16 )
2x2 –16x +16 +k
=k
=k
=k
=0
Since the equation
2x2 –16x +16 + k
= 0 has real roots ,
the discriminant D  0
a=2
b = -16
c = 16+k
D = b2 – 4ac 0
i.e
( -16)2 – 4 ( 2 ) ( 16 +k )
256 –128 -8k
128 -8k
-8k
k
0
0
0
 -128
16
13.Given that x(x-2) = t – 2 , find x in terms of t. Hence , find
the range of values of t for x to be real.
The Given equation can be written as
x2 – 2x –t +2 = 0
Since
a = 1, b = - 2 , c = 2 – t
D
= b2 – 4ac
=( -2 )2 – 4 ( 1) ( 2- t)
=4 – 8 +4t
=-4+4t
x = 1 sqr (t-1)
x  1  t 1
For the equation x2 – 2x –t +2 = 0
has real roots , the discriminant D  0
i.e
-4+4t
4t
t
0
4
1