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Quadratic Equations - Solved Subjective Questions Part 2

Posted on - 10-05-2017

JEE Math QE

IIT JEE

Example 1

Show that the value of , whenever defined, never lies between 1/3 and 3.

Solution:

Since tan2x is positive, either y > 3 or y < 1/3. If y lies between 1/3 and 3 i.e. if 1/3 < y < 3, tan2x is negative, which is not possible. .

Hence never lies between 1/3 and 3.

Example 2

If x1, x2 are the roots of x2 - 3x + a = 0 and x3, x4 are those of the equation x2 - 12x + b = 0 and x1, x2, x3, x4 are in increasing G.P., find a and b.

Solution:

Let x1 = A and r be the common ratio. We have.

x1 = A, x2 = Ar, x3 = Ar2, x4 = Ar3

also, x1 + x2 = 3 …(1)

x1 x2 = a …(2)

and x3 + x4 = 12 …(3)

x3 x4 = b …(4)

From equation (1) and (3), we have
&⇒
r = ± 2


&⇒
r = 2 (since given sequence is an increasing G
.P.).

Also A (1+ r) = 3
&⇒
A = 1

Therefore, x1=1, x2 = 2, x3= 4, x4 = 8

Thus we have a = x1x2 = 2, b = x3x4= 32

Example 3

If x2 + cx + ab = 0 and x2 + bx + ca = 0 have a common root, then show that their other roots satisfy the equation x2 + ax + bc = 0.

Solution:

Let the equation x2 + cx + ab = 0 …(1)

and x2 + bx + ca = 0 …(2)

have the common root a

Therefore, a2 + ca + ab = 0

And a2 + ba + ca = 0


&⇒


&⇒
a = - (b + c) and also a = a
&⇒
a + b + c = 0

Let the other root of equations (1) and (2) be b and b1 respectively then a×b = ab and ab1 = ca

But a = a. Therefore b = b, b1 = c.

Now the equation whose roots are b and b1 is

x2 – (b + b1)x + b×b1 = 0
&⇒
x2 – (b + c)x + bc = 0


&⇒
x2 + ax + bc = 0 (Since b + c = -a)

Hence other roots of equation (1) and (2) are given by the equation

x2+ ax + bc = 0

Example 4

If 2x3 + ax2 +bx +4 = 0 ( a and b are positive real numbers ) has 3 real roots, then prove that a+ b ≥ 6(21/3 + 41/3). .

Solution:

Let a, b, g be the roots of 2x3 + ax2 + bx + 4 = 0. Given that all the coefficients are positive, so all the roots will be negative.

Let a1 = -a, b2 = -b, a3 = – r


&⇒
a1 + a2 + a3 =

a1 a2 + a2a3 + a3a1=

a1 a2 a3 = 2

Applying A.M. ≥ G.M. , we have.

≥ (a1 a2a3)1/3
&⇒
≥ 2
&⇒
a ≥ 6 ´ 21/3

Also ≥ (a1 a2 a3)2/.3


&⇒
≥ 4
&⇒
b ≥ 6 ´41/3

Therefore a + b ≥ 6 (21/3 + 41/3) = 6(21/3 + 41/3)

Example 5

Show that the equation = k has no imaginary root, where A, B, C ……,H and a, b, c ….., h and k ∈ R

Solution:

Suppose one root of the equation is (u +iv) then other root would be u – iv


&⇒
….(1)

and ….(2)

(1) –(2) , we get

iv

This is possible only when v = 0 and for this case there is no imaginary root.

 
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