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1

### AIEEE 2005

In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then
A
$$a = b + c$$
B
$$c = a + b$$
C
$$b = c$$
D
$$b = a + c$$

## Explanation

$$\angle$$R = 90o $$\therefore$$ $$\angle$$P + $$\angle$$Q = 90o

$$\Rightarrow$$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$o

$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$

$$\therefore$$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$

and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$

$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$

$$= \tan \left( {{P \over 2} + {Q \over 2}} \right)$$= tan 45o = 1

$$\Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$

$$\Rightarrow - {b \over a} = {a \over a} - {c \over a}$$

$$\Rightarrow - b = a - c$$ or $$c = a + b.$$
2

### 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.

$$\therefore$$ $${p^2} - 4q = 0$$ $$\Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$
3

### 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 $$p-1=0$$
4

### 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

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

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

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