QUESTION

# If a $\in$ (-1, 1), then roots of the quadratic equation $\left( a-1 \right){{x}^{2}}+ax+\sqrt{1-{{a}^{2}}}=0$a) Realb) Imaginaryc) Both equald) None of these

Hint: First write the equation for discriminant by using the formula $D={{b}^{2}}-4ac$ and then use the condition a $\in$ (-1, 1) and adjust it with the terms of ‘D’ to define whether ‘D’ is +ve, -ve or zero and use the conditions of discriminant to get the final answer.

To solve the above question we will write the given data first, therefore,

a $\in$ (-1, 1) ……………………………………………………………………….. (1)

$\left( a-1 \right){{x}^{2}}+ax+\sqrt{1-{{a}^{2}}}=0$

If we compare the above equation with $a{{x}^{2}}+bx+c=0$ we will get,

a = $\left( a-1 \right)$ , b = a, c = $\sqrt{1-{{a}^{2}}}$ ………………………………………………………… (2)

As we have given a quadratic equation and all the options are asking for the type of solution therefore we only have to deal with the Discriminant and for that we should know the formula given below,

Formula:

$D={{b}^{2}}-4ac$

If we put the values of equation (1) in the above formula we will get,

$D={{a}^{2}}-4\left( a-1 \right)\sqrt{1-{{a}^{2}}}$ ……………………………………………………………….. (3)

Now consider equation (1),

a $\in$ (-1, 1)

This can also be written as,

-1 < a < 1

If we square above inequation we will get the value of ‘a’ in between 0 and 1 as the square of negative part of the inequation will become positive therefore we can write,

$0<\text{ }{{a}^{2}}\text{ }<\text{ }1$

Multiplying by -1 in the above inequation we will get,

$0>\text{ }-\,{{a}^{2}}\text{ }-\text{1}$ Therefore we can write,

${{a}^{2}}\in \text{(0,1)}$ i.e. +ve value ………………………………………………….. (4)

If we add 1 in the above inequation we will get,

$1+0>\text{ 1}-{{a}^{2}}\text{ 1}-\text{1}$

By simplification in the above inequation we will get,

$1>\text{ 1}-{{a}^{2}}\text{ 0}$Therefore we can write,

$\text{1}-{{a}^{2}}\in \text{(0,1)}$ i.e. +ve value …………………………………………………… (5)

Also, if we subtract 1 from -1 < a < 1 we will get,

$-1-1<\text{ }a-1\text{ }<\text{ }1-1$

Therefore, $-2<\text{ }a-1\text{ }<\text{ 0}$

Therefore we can write,

$a-1\in \text{(}-\text{2,0)}$ i.e. -ve value …………………………………………………… (6)

If we put the values of equation (4), equation (5) and equation (6) in terms of sign in

equation (3) we will get,

Therefore, D = (+ve value) – 4(-ve value)(+ve value)

If we take -1 out from –ve value then it will become +ve value therefore above equation will become,

Therefore, D = (+ve value) – 4(-1)(+ve value)(+ve value)

Therefore, D = (+ve value) + 4(+ve value)(+ve value)

Now if we observe the above equation then we can say that the overall value of discriminant is +ve,

Therefore, D = +ve

Therefore, D > 0 …………………………………………………….. (7)

Now to state the type of solution we will get we should know the conditions given below,
Conditions:

If D > 0, - Solution is real and has distinct roots.

If D = 0, - Solution is real and saves equal roots.

If D < 0, - Solution is imaginary.

If we compare equation (7) with above equations then we can say that, given equations have real roots.

Therefore the correct answer is option (a)

Note: Do not solve the equation $D={{a}^{2}}-4\left( a-1 \right)\sqrt{1-{{a}^{2}}}$ further as it will complicate your answer. In place of that use the given condition to find whether D is +ve, -ve, or equal to zero.