Question

The roots of the equations ${x^2} - x - 3 = 0$ are?

Answer 1

The given equation is to solve for the roots of the quadratic equation, for which we need to use the sridharacharya rule. This rule is used to directly state the roots of the quadratic equation, here we need to first find the determinant “d” then we can write the roots for the equation.
Formulae Used: For the roots of the quadratic equation, of general equation say:
$\Rightarrow a{x^2} + bx + c = 0$
Determinant “d” can be written as:
$\Rightarrow d = \sqrt {{b^2} - 4ac}$
Roots of the equations are:
$\Rightarrow roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
Here the given question needs to find the roots for the quadratic equation, for which we need to first find the determinant of the equation, on solving we get:
$\Rightarrow d = \sqrt {{b^2} - 4ac} = \sqrt {{1^2} - 4 \times 1 \times ( - 3)} = \sqrt {13}$
Now roots of the equation can be written as:
$\Rightarrow roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - 1 \pm \sqrt {13} }}{{2 \times 1}} = \dfrac{{ - 1 \pm \sqrt {13} }}{2}$
Here we got the roots for the given quadratic equation.
Additional Information: Here the determinant is non negative hence real determinant is obtained, but if the determinant comes to be negative then we have to solve it by using complex number properties.

Note: The given question needs to be solved by using the sridharacharya rule, here we cannot use mid term splitting rule because the middle term cannot be broken or split according to the mid term split rule.