Question
Question: A quadratic equation is given as \(a{{x}^{2}}+bx+c=0\) , where a,b,c are real, has real roots if: ...
A quadratic equation is given as ax2+bx+c=0 , where a,b,c are real, has real roots if:
(A). a , b, c are integers
(B). b2>3ac
(C). ac>0
(D). c=0
Solution
Hint: A quadratic equation has 2 roots. Take them as α and β .Find the discriminant (D) from this. For different cases where D>0,D=0 and D<0 , find which of the above gives real root.
Complete step-by-step solution -
We know that a quadratic equation has 2 roots. Now let α and β be the roots of the general form of the quadratic equation ax2+bx+c=0 .Then we can write,
α=2a−b−b2−4ac and β=2a−b+b2−4ac .
We have been told that a, b and c are real. Hence the nature of the roots α and β of equation ax2+bx+c=0 depends on the quantity (b2−4ac)which is called the discriminant of the quadratic equation ax2+bx+c=0.
We can consider different cases where a, b and c are real and a=0 .
Case 1: b2−4ac>0 , discriminant (D) is positive and has real and unequal roots.
Case 2: b2−4ac=0 , D is zero and has real and equal roots.
Case 3: b2−4ac<0 , D is negative and has roots which are imaginary.
From these 3 cases, we can say that b2−4ac>0 , the roots are real and for b2−4ac=0 , the roots are real and equal. Hence if we put c=0 , we get b2 which will always be positive.
If we put ac>0 , but if it is greater than b2 then the roots will be imaginary.
Thus, a, b, c are real and have real roots if c=0.
∴c=0 is the correct answer.
∴ Option (D) is the correct answer.
Note: There are other cases for (b2−4ac) when it is a perfect square and not perfect square. But those cases are not relevant here. Thus, when the determinant of the equation should be positive for its roots to be real is b2−4ac>0 . But from the option c=0 is correct. As it reduces determinant b2 which is always positive.