# For a quadratic equation if D < 0 then which of the following is true?

(a) Real roots do not exist

(b) Roots are real and equal

(c) Roots are rational and distinct

(d) Roots are real and distinct

Last updated date: 29th Mar 2023

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Answer

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Hint: Recollect the quadratic formula which is used to find the roots of a quadratic equation. This quadratic formula consists of a term which is discriminant in it. Assume the discriminant < 0 and try to find the nature of roots of the quadratic equation.

Before proceeding with the question, we must know the formulas that will be required to solve this question.

In quadratic equations, we have quadratic formula from which, the roots of the equation $a{{x}^{2}}+bx+c=0$ are given by,

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

In this formula, the term ${{b}^{2}}-4ac$ is the discriminant because discriminant $D={{b}^{2}}-4ac$. Substituting ${{b}^{2}}-4ac=D$ in the above equation, we get,

$x=\dfrac{-b\pm \sqrt{D}}{2a}...................\left( 1 \right)$

From the above equation, we can find the nature of roots for different values of discriminant $D$.

In the question, it is given that the discriminant $D$ for a quadratic equation is less than 0 and we have to find the nature of the roots.

Since it is given that the discriminant is less than 0, let us put substitute D<0 in equation $\left( 1 \right)$. From the complex numbers, we know that the square root of a negative number is an imaginary number. In equation $\left( 1 \right)$, D is negative and it is inside the square root. So, we can say that x is an imaginary number i.e. x is not a real number. So, the real roots of the quadratic equation do not exist.

Hence, the answer is option (a).

Note: To solve this question in a lesser time, one should remember it as a fact that any quadratic equation with discriminant less than 0 does not have real roots. One should also remember that for discriminant = 0, the quadratic equation has real and equal roots and for discriminant greater than 0, the quadratic equation has real and distinct roots.

Before proceeding with the question, we must know the formulas that will be required to solve this question.

In quadratic equations, we have quadratic formula from which, the roots of the equation $a{{x}^{2}}+bx+c=0$ are given by,

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

In this formula, the term ${{b}^{2}}-4ac$ is the discriminant because discriminant $D={{b}^{2}}-4ac$. Substituting ${{b}^{2}}-4ac=D$ in the above equation, we get,

$x=\dfrac{-b\pm \sqrt{D}}{2a}...................\left( 1 \right)$

From the above equation, we can find the nature of roots for different values of discriminant $D$.

In the question, it is given that the discriminant $D$ for a quadratic equation is less than 0 and we have to find the nature of the roots.

Since it is given that the discriminant is less than 0, let us put substitute D<0 in equation $\left( 1 \right)$. From the complex numbers, we know that the square root of a negative number is an imaginary number. In equation $\left( 1 \right)$, D is negative and it is inside the square root. So, we can say that x is an imaginary number i.e. x is not a real number. So, the real roots of the quadratic equation do not exist.

Hence, the answer is option (a).

Note: To solve this question in a lesser time, one should remember it as a fact that any quadratic equation with discriminant less than 0 does not have real roots. One should also remember that for discriminant = 0, the quadratic equation has real and equal roots and for discriminant greater than 0, the quadratic equation has real and distinct roots.

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