if b2−4ac≥0, then write the roots of a quadratic equation ax2+bx+c=0.

- b + b 2 - 4 ac 2 a and - b - b 2 - 4 ac 2 a

if b2−4ac≥0, then write the roots of a quadratic equation ax2+bx+c=0.

- b + b 2 - 4 ac 2 a and - b - b 2 - 4 ac 2 a

- b + b 2 + 4 ac 2 a and - b - b 2 + 4 ac 2 a

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if $b^{2}$ −4ac $\geq$ 0, then write the roots of a quadratic equation $a x^{2} + bx + c = 0$ .

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

and

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

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

and

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

0 and 0

0 and 1

answer is A.

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Detailed Solution

Step1
Given equation

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

is in one variable x in which a≠0&b,c⩾0.
A is the coefficient of

x^{2}

, b is the coefficient of x and c is the constant term. Here, a, b and c are the real numbers.
We have to find the roots of the given equation

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

. i.e. we have to find the two values of x. when we put the value of x in the given equation, it becomes zero.
In the given equation

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

only one condition is given

b^{2}

−4ac

\geq

0,
Step 2
Consider the quadratic equation

{ax}^{2} + bx + c = 0

where a≠0.
Dividing throughout both sides by a , we get

x^{2}

\frac{b}{a}

\frac{c}{a}

=0. Here in first terms a cancel by the coefficient of

x^{2}

. So, using the completing square method. take half of the coefficient of x then add and subtract the squares in the equations.

x^{2} \frac{b}{a}

x +

{(\frac{b}{2 a})}^{2}

−

{(\frac{b}{2 a})}^{2}

\frac{c}{a}

=0.now, we use the identity

{(a + b)}^{2} = a^{2} + b^{2} + 2 ab

We get,

{(x + \frac{b}{2 a})}^{2} - \frac{b^{2}}{4 a^{2}} + \frac{c}{a}

=0
This is the same as

{(x + \frac{b}{2 a})}^{2} - \frac{b^{2} - 4 ac}{4 a^{2}}

=0
Take the second term to the right-hand side of the equation we get;

{(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 ac}{4 a^{2}}

So, the roots of the given equation are the same as those of

{(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 ac}{4 a^{2}}

Step 3
If

b^{2}

−4ac

\geq

0 , then by taking the square roots, we get

(x + \frac{b}{2 a})

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

Therefore, x=−

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

by lcm, we get x=

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

So, the roots of the equation are x=

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

and x=

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

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