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Quadratic Equations - Solved Subjective Questions Part 3

Posted on - 08-05-2017

JEE Math QE

IIT JEE

Example 1

If a, b are the roots of x2+px+q=0, and g,d are the roots of x2 + 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.

Solution:

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

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

= [a2-a(g+d)+ gd][b2- b(g+d)+ gd]

= [a2+ra+s][b2+rb+s] = a2b2+rab(a+b)+r2ab+s[a2+b2]+s2+sr(a+b)

= q2-rpq + r2q + s(p2-2q)+s2-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.

Example 2

Find all the real numbers x satisfying

Solution:

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.

Example 3

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.

Solution:

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, y2 +y(x-6) + x2 –6x +7 =0

Since y is real , (x-6)2 - 4(x2 –6x + 7 ) ≥ 0

or, 3x2 -€‘ 12x –8 £ 0
&⇒

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

Example 4

If one root of the quadratic equation ax2 + bx + c = 0 is equal to nth ( n is an even natural number) power of the other root, then show that .

Solution:

Let a be one of the roots. Then the other root is an.

Hence a+an = -b/a and an+1 = c/a

.

Example 5

Find all the values of the parameter c for which the inequality
1 + log2(2x2 +2x +) ≥ log2(cx2 +c) possesses at least one solution.

Solution:

Given inequality is 1 + log2(2x2 +2x +) ≥ log2(cx2 +c) ….. (1)

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

2x2 + 2x += 2(x2 + x + ) = 2" x ∈ R

We can write (1) as log2


&⇒
cx2 + c £ 4x2+ 4x + 7


&⇒
(c - 4)x2 - 4x +c - 7£ 0 …
..(2).

We have D = b2 - 4ac = 16 - 4(c - 4)(c - 7)

= -4(c2 - 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)

 
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