If a, b
are the roots of x^{2}+px+q=0, and g,d are the roots of x^{2 }+ rx
+ s = 0, evaluate (a – g) (a – d)
(b – g) (b – d)
in terms of p, q, r and s.

Deduce the condition that the equations have a common root.

We have a+b=-p, ab=q, g+d= - r, gd=s

Hence (a-g) (a-d) (b-g) (b-d)

= [a^{2}-a(g+d)+
gd][b^{2}- b(g+d)+
gd]

= [a^{2}+ra+s][b^{2}+rb+s] = a^{2}b^{2}+rab(a+b)+r^{2}ab+s[a^{2}+b^{2}]+s^{2}+sr(a+b)

= q^{2}-rpq + r^{2}q
+ s(p^{2}-2q)+s^{2}-srp = (q-s)^{2} + (p-r) (sp-rq)

For a common root
R.H.S. = 0

&⇒ (q-s)^{2} = (p-r) (rq-sp),
which is the required condition.

Find all the real numbers x satisfying

Let= A

and = B

We have A + B = and AB =

Therefore (A - B)^{2} = (A +
B)^{2} - 4AB

= 4. - 4. = 0

&⇒ A = B

Since powers in A and B are same, either the power is equal to zero or the bases are the same.

&⇒
x = 0 or

&⇒
x = 0 or =0

&⇒ x = 0 or x
= 4 or 1.

Let
x, y, z be real variables satisfying the equations x+y +z = 6 and
xy + yz +zx = 7. Then find the range in which the variables

can lie.

We are given that

x+ y+ z = 6 . . . . (1).

and xy +yz +zx =7 . . . . . (2) .

From (1), z = 6 - x -y

Putting the value of z in (2), we get

xy+ y(6-x-y) + x( 6-x-y) =7

or, y^{2} +y(x-6) + x^{2}
–6x +7 =0

Since y is real , (x-6)^{2}
- 4(x^{2} –6x + 7 ) ≥ 0

or, 3x^{2 }- 12x –8 £
0

&⇒

Since (1) and (2) are symmetrical in x, y and z, all the variable lie in the interval .

If one root of the
quadratic equation ax^{2} + bx + c = 0 is equal to nth ( n is an even
natural number) power of the other root, then show that .

Let a be one of the
roots. Then the other root is a^{n}.

Hence a+a^{n} = -b/a and a^{n+1
}= c/a

.

Find all the
values of the parameter c for which the inequality

1 + log_{2}(2x^{2} +2x +) ≥ log_{2}(cx^{2}
+c) possesses at least one solution.

Given inequality is 1 + log_{2}(2x^{2}
+2x +) ≥ log_{2}(cx^{2}
+c) ….. (1)

In order that (1) make sense we must have c> 0. Also .

2x^{2} + 2x += 2(x^{2} + x + ) = 2" x ∈
R

We can write (1) as log_{2}_{}

&⇒
cx^{2} + c £ 4x^{2}+ 4x + 7

&⇒
(c - 4)x^{2} - 4x +c - 7£ 0 …..(2).

We have D = b^{2} - 4ac = 16
- 4(c - 4)(c - 7)

= -4(c^{2} - 11c + 24) = -4(c
- 3)(c - 8)

Case-I : If D < 0, that is, if
c < 3 or c > 8 and c - 4< 0 then(2) holds for all x∈
R. Thus if 0 < c < 3. Then (1) holds for

all x ∈ R. .

Case-II: If D ≥ 0, that is, if 3 £ c £ 8 and c – 4< 0

then (2) holds for all x £ -(4 + )/2(4 - c) and for all

x ≥ -(4 - )/2(4 - c). Thus,

if 3 £ c < 4 then there is at least one x satisfying (1). .

Case-III: If D ≥ 0, that is, if 3 £ c £ 8 and c - 4 > 0 then (2) holds for all x lying between -(4+)/2(4 - c) and -(4 +)/2(4 -c). Thus for 4 < c £ 8 there is at least one x satisfying (1). Also for c = 4, (2) is satisfied for all x ≥ -3/4

Hence if 0 < c £ 8, there is at least one x satisfying the inequality (1)