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1

### AIEEE 2004

If one root of the equation $${x^2} + px + 12 = 0$$ is 4, while the equation $${x^2} + px + q = 0$$ has equal roots,
then the value of $$'q'$$ is
A
4
B
12
C
3
D
$${{49} \over 4}$$

## Explanation

$$4$$ is a root of $${x^2} + px + 12 = 0$$

$$\Rightarrow 16 + 4p + 12 = 0$$

$$\Rightarrow p = - 7$$

Now, the equation $${x^2} + px + q = 0$$

has equal roots.

$$\therefore$$ $${p^2} - 4q = 0$$ $$\Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$
2

### AIEEE 2004

If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are
A
$$- 1,2$$
B
$$- 1,1$$
C
$$0,-1$$
D
$$0,1$$

## Explanation

Let the second root be $$\alpha .$$

Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$

Also $$\alpha .\left( {1 - p} \right) = 1 - p$$

$$\Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$

$$\Rightarrow p = 1$$ [as $$\alpha = - 1$$]

$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$
3

### AIEEE 2004

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation
A
$${x^2} - 18x - 16 = 0$$
B
$${x^2} - 18x + 16 = 0$$
C
$${x^2} + 18x - 16 = 0$$
D
$${x^2} + 18x + 16 = 0$$

## Explanation

Let two numbers be a and b then $${{a + b} \over 2} = 9$$

and $$\sqrt {ab} = 4$$

$$\therefore$$ Equation with roots $$a$$ and $$b$$ is

$${x^2} - \left( {a + b} \right)x + ab = 0$$

$$\Rightarrow {x^2} - 18x + 16 = 0$$
4

### AIEEE 2003

The real number $$x$$ when added to its inverse gives the minimum value of the sum at $$x$$ equal to
A
-2
B
2
C
1
D
-1

## Explanation

$$y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$

For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$

$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} \Rightarrow {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 2}} = 2$$ ($$+ve$$ minima)

$$\therefore$$ $$x=1$$

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