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Q.

If $b^{2} - 4 ac \geq 0$ then the roots of quadratic equation $a x^{2} + bx + c = 0$ is

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a

\frac{b}{2 a} + \frac{\sqrt{b^{2} - 4 ac}}{2 a}

b

- \frac{b}{2 a} - \frac{\sqrt{b^{2} + 4 ac}}{2 a}

c

\frac{b}{2 a} \pm \frac{\sqrt{b^{2} + 4 ac}}{2 a}

d

- \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 ac}}{2 a}

answer is D.

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Detailed Solution

It is given that for the equation

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

,

b^{2} - 4 ac \geq 0 .

Here are the relations between roots and discriminant
●       When roots are non-real, the discriminant is less than 0
●       When roots are real and equal, the discriminant is equal to 0
●       When the roots real and unequal the discriminant is greater than 0
If

b^{2} - 4 ac \geq 0

then the roots are real and distinct,
Let the roots be α and β.
We know that,
α+β=

- \frac{b}{a}

…..(1)
αβ=

\frac{c}{a} \dots

..(2)
We can write that,

{(α - β)}^{2} = {(α + β)}^{2} - 4 αβ

Substituting the values of α+β and αβ in the above equation,

α - β = \frac{b^{2} - 4 ac}{a}

…… (3)
Adding equation 1 and 3, we get,

α = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

Hence, by using quadratic formula, we get the roots as

\frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

Hence, option 4 is correct.

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