# Let $a,{\text{ }}b$ and $c$ be the real numbers such that $4a + 2b + c = 0 $ and $ ab > 0$. Then the equation $a{x^2} + bx + c = 0$ has:

$(a)$complex roots

$(b)$exactly one root

$(c)$real roots

$(d)$none of these

Answer

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Hint: In the above given question, we are asked to calculate the nature of the roots of the quadratic equation $a{x^2} + bx + c = 0$. And we must be aware of the fact that in the polynomial equations, the number of roots is equal to its degree.

We have the given quadratic equation as

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

Let us assume the value $x = 2$ in the above equation, we get

$4a + 2b + c = 0$

which means that one real root of the given quadratic equation can be equal to 2.

As we already know that the number of roots of a polynomial equation is equal to its degree, therefore, we can conclude that the given quadratic equation $a{x^2} + bx + c = 0$ also has two roots.

Now, since one real root of the quadratic equation is 2, therefore, the other root must also be real.

Hence, the roots of the given quadratic equation are real and distinct.

So, the required solution is the option $(c)$ real roots.

Note: Whenever we face such types of problems the key point is to have a good grasp of the nature of the roots like real and distinct, real and equal, imaginary roots, etc. After determining the nature of roots of the given polynomial equation, the value of the roots can be easily calculated.

We have the given quadratic equation as

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

Let us assume the value $x = 2$ in the above equation, we get

$4a + 2b + c = 0$

which means that one real root of the given quadratic equation can be equal to 2.

As we already know that the number of roots of a polynomial equation is equal to its degree, therefore, we can conclude that the given quadratic equation $a{x^2} + bx + c = 0$ also has two roots.

Now, since one real root of the quadratic equation is 2, therefore, the other root must also be real.

Hence, the roots of the given quadratic equation are real and distinct.

So, the required solution is the option $(c)$ real roots.

Note: Whenever we face such types of problems the key point is to have a good grasp of the nature of the roots like real and distinct, real and equal, imaginary roots, etc. After determining the nature of roots of the given polynomial equation, the value of the roots can be easily calculated.

Last updated date: 18th Sep 2023

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