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Q.

if b2−4ac0, then write the roots of a quadratic equation ax2+bx+c=0.


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a

-b+b2-4ac2a and -b-b2-4ac2a

b

-b+b2+4ac2a and -b-b2+4ac2a

c

0 and 0

d

0 and 1 

answer is A.

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Detailed Solution

Step1
Given equation ax2+bx+c=0 is in one variable x in which a≠0&b,c0.
A is the coefficient of x2  , 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 ax2+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 ax2+bx+c=0  only one condition is given b2−4ac0,
Step 2
Consider the quadratic equation ax2+bx+c=0   where a≠0.
Dividing throughout both sides by a , we get x2+bax+ca=0. Here in first terms a cancel by the coefficient of x2. So, using the completing square method. take half of the coefficient of x then add and subtract the squares in the equations.
 x2bax + (b2a)2(b2a)2+ca=0.now, we use the identity (a+b)2=a2+b2+2ab
We get, (x+b2a)2-b24a2+ca=0
This is the same as (x+b2a)2-b2-4ac4a2=0
Take the second term to the right-hand side of the equation we get;
(x+b2a)2=b2-4ac4a2
 So, the roots of the given equation are the same as those of (x+b2a)2=b2-4ac4a2
Step 3
If b2−4ac0 , then by taking the square roots, we get (x+b2a)=b2-4ac2a
Therefore, x=−b2a+b2-4ac2a by lcm, we get  x=-b±b2-4ac2a
So, the roots of the equation are x=-b+b2-4ac2a  and x=-b-b2-4ac2a
 
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