Question

write the roots of the quadratic equation X square + bx + c is equal to zero , where a is not equal to zero​

Answer 1

Step-by-step explanation:

ax^2+bx+c=0

x={-b±√(b^2–4ac)}/2a

as per condition

{-b+√(b^2–4ac)}/2a={-b-√(b^2–4ac)}/2a

or, -b+√(b^2–4ac)=-b-√(b^2–4ac)

or, √(b^2–4ac)=-√(b^2–4ac)

or, 2√(b^2–4ac)=0

or b^2=4ac

now x ={-b+√(b^2–4ac)}/2a

={-b+√(b^2-b^2)}2a

=-b/2a

equal root is -b/2a

Hope it helps..

Answer 2

Answer:

The roots of the given quadratic equation are $\frac{-b + \sqrt{b^{2}- 4c } }{2} and \frac{-b -\sqrt{b^{2}- 4c } }{2}$

Step-by-step explanation:

Explanation:

Given , $x^{2} +bx +c = 0$

Discriminant - The discriminant is  part of the quadratic equation under the square root symbol b²- 4ac. The discriminant indicates whether there are two solutions, one solution, or no solution.

Step 1:

We have , $x^{2} +bx +c = 0$

From discriminant formula we have ,

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

Now , here we see that the coefficient of $x^{2}$ is 1  and coefficient of 'x' is b and constant is 'c ' .

Now put the value in the discriminant  formula ,

x = $\frac{-b \frac{+}{} \sqrt{b^{2}- 4.1.c } }{2.(1)}$ = $\frac{-b \frac{+}{} \sqrt{b^{2}- 4c } }{2}$

x = $\frac{-b + \sqrt{b^{2}- 4c } }{2} , \frac{-b -\sqrt{b^{2}- 4c } }{2}$

Final answer:

Hence , $\frac{-b + \sqrt{b^{2}- 4c } }{2} and \frac{-b -\sqrt{b^{2}- 4c } }{2}$ are the roots of the given quadratic equation.

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