Question

The root $a{x^2} + x + 1 = 0$, where $a \ne 0$are in the ratio 1: 1. The a is equal to
A.$\dfrac{1}{4}$
B.$\dfrac{1}{2}$
C.$\dfrac{3}{4}$
D.1

Answer 1

Hint: A general quadratic equation can be written as $a{x^2} + bx + c = 0$
Roots of the above quadratic equation can be calculated as : $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ which is called Discriminant method.

Complete step-by-step answer:
 The roots of the given equation $a{x^2} + x + 1 = 0$ are given in the ratio 1: 1. Where $a \ne 0$.
We are to find the value of a:-
Now, we know that, if the roots of quadratic equation are in 1 : 1 that means the roots are equal and if the roots are equal, then
Discriminant, D = 0
 $ \Rightarrow {b^2} - 4ac = 0$
$ \Rightarrow {b^2} = 4ac................(i)$
In the given equation , a =a , b =1 and c =1
So equation (i), becomes
$ \Rightarrow {1^2} = 4a(1)$
$ \Rightarrow 1 = 4a$
$ \Rightarrow a = \dfrac{1}{4}$
Hence, option (A) is correct a= $\dfrac{1}{4}$.

Note: The roots of the quadratic equation $a{x^2} + bx + c = 0$ can be calculated with the formula :
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where D = ${b^2} - 4ac$is called Discriminant.
If D>0 two distinct real roots would occur.
If D<0, no real roots would be there.
If D = 0, two real roots would be there as in the given problem.
Hence, by comparing the given equation $a{x^2} + x + 1 = 0$ with the general equation,
We get a= a, b =1 and c =1. So, by using the formula D=0,
 $ \to {b^2} - 4ac = 0$