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1

### WB JEE 2010

If $$\alpha$$, $$\beta$$ be the roots of the quadratic equation x2 + x + 1 = 0 then the equation whose roots are $$\alpha$$19, $$\beta$$7 is

A
$${x^2} - x + 1 = 0$$
B
$${x^2} - x - 1 = 0$$
C
$${x^2} + x - 1 = 0$$
D
$${x^2} + x + 1 = 0$$

## Explanation

$$\because$$ $$\alpha$$, $$\beta$$ are roots of $${x^2} + x + 1 = 0$$ (given)

But $$\omega$$, $$\omega$$2 are roots of $${x^2} + x + 1 = 0$$

$$\therefore$$ $$\alpha$$ = $$\omega$$, $$\beta$$ = $$\omega$$2 (say)

$$\therefore$$ $${\alpha ^{19}} + {\beta ^7} = {\omega ^{19}} + {\omega ^{14}} = \omega + {\omega ^2} = - 1$$ [$$\because$$ $${\omega ^3} = 1$$ & $$1 + \omega + {\omega ^2} = 0$$]

$${\alpha ^{19}}\,.\,{\beta ^7} = {\omega ^{19}}\,.\,{\omega ^{14}} = {\omega ^{33}} = 1$$

$$\therefore$$ Required equation is $${x^2} - ( - 1)x + 1 = 0$$

i.e. $${x^2} + x + 1 = 0$$

2

### WB JEE 2009

If a, b, c are real, then both the roots of the equation $$(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b) = 0$$ are always

A
positive
B
negative
C
real
D
imaginary

## Explanation

Given equation is

$$(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b) = 0$$

or, $$3{x^2} - 2(a + b + c)x + ab + bc + ca = 0$$

$$D = {b^2} - 4ac = 4{(a + b + c)^2} - 4\,.\,3(ab + bc + ca)$$

$$= 4[{a^2} + {b^2} + {c^2} - ab - bc - ca]$$

$$= 2[{(a - b)^2} + {(b - c)^2} + {(c - a)^2}] \ge 0$$

So roots are real.

3

### WB JEE 2009

The quadratic equation whose roots are three times the roots of $$3a{x^2} + 3bx + c = 0$$ is

A
ax2 + 3bx + 3c = 0
B
ax2 + 3bx + c = 0
C
9ax2 + 9bx + c = 0
D
ax2 + bx + 3c = 0

## Explanation

$$3a{x^2} + 3bx + c = 0$$

Let roots of given equation are $$\alpha$$, $$\beta$$

$$\therefore$$ $$\alpha + \beta = - {b \over a}$$

$$\alpha \beta = {c \over {3a}}$$

$$\therefore$$ New roots are $$\gamma$$, $$\delta$$ $$\therefore$$ $$\gamma$$ = 3$$\alpha$$, $$\delta$$ = 3$$\beta$$

$$\gamma + \delta = 3(\alpha + \beta ) = - {{3b} \over a}$$

$$\gamma \delta = 3\alpha \,.\,3\beta = {{3c} \over a}$$

$$\therefore$$ Quadratic equation is $${x^2} - (\gamma + \delta )x + \gamma \delta = 0$$

$$\Rightarrow {x^2} + {{3b} \over a}x + {{3c} \over a} = 0 \Rightarrow a{x^2} + 3bx + 3c = 0$$

4

### WB JEE 2009

The sum of all real roots of the equation $$|x - 2{|^2} + |x - 2| - 2 = 0$$ is

A
7
B
4
C
1
D
5

## Explanation

$$|x - 2{|^2} + |x - 2| - 2 = 0$$

Let $$|x - 2| = t$$

$$\therefore$$ $${t^2} + t - 2 = 0 \Rightarrow (t + 2)(t - 1) = 0$$

$$\Rightarrow$$ t = $$-$$2 or 1

If t = $$-$$2 then $$|x - 2| = - 2$$, but modulus of any value is non negative, so it has no solution.

If t = 1, then $$|x - 2| = 1 \Rightarrow x - 2 = \pm 1 \Rightarrow x = 3$$ or 1

$$\therefore$$ Sum of roots = 3 + 1 = 4

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