Question

Find the nature of the roots of the following quadratic equations. If the real roots exists, find them:
(i). $2{x^2} - 3x + 5 = 0$
(ii). $3{x^2} - 4\sqrt 3 x + 4 = 0$
(iii). $2{x^2} - 6x + 3 = 0$

Answer 1

We have found the nature of the roots of the given quadratic equations. By using the formula,
The roots of a quadratic equation $a{x^2} + bx + c = 0$ is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The type of the roots of the quadratic equation is calculated by its discriminant ${b^2} - 4ac$.
(1). ${b^2} - 4ac > 0$, then two roots are different real numbers;
(2). ${b^2} - 4ac = 0$, then two roots are equal real numbers;
(3). ${b^2} - 4ac < 0$, then two roots are imaginary numbers.

Complete step-by-step answer:
(i). The given equation is $2{x^2} - 3x + 5 = 0$.
We need to find out the nature of the roots of the given quadratic equation.
Comparing this equation with $a{x^2} + bx + c = 0$
We get, $a = 2,b = - 3,c = 5$
Thus, ${b^2} - 4ac = {( - 3)^2} - 4 \times 2 \times 5 = 9 - 40 = - 31 < 0$
Thus the two roots are imaginary numbers.

(ii). The given equation is $3{x^2} - 4\sqrt 3 x + 4 = 0$.
We need to find out the nature of the roots of the given quadratic equation.
Comparing this equation with $a{x^2} + bx + c = 0$
We get, $a = 3,b = - 4\sqrt 3 ,c = 4$
Thus, ${b^2} - 4ac = {( - 4\sqrt 3 )^2} - 4 \times 3 \times 4 = 48 - 48 = 0$
Thus the two roots are equal real numbers.
Now we will apply Sri dharacharya’s formula.
The roots are given by,
$x = \dfrac{{ - ( - 4\sqrt 3 ) \pm 0}}{{2 \times 3}} = \dfrac{{4\sqrt 3 }}{6} = \dfrac{2}{{\sqrt 3 }}$
Thus the real and equal roots are $\dfrac{2}{{\sqrt 3 }},\dfrac{2}{{\sqrt 3 }}$.

(iii). The given equation is $2{x^2} - 6x + 3 = 0$.
We need to find out the nature of the roots of the given quadratic equation.
Comparing this equation with $a{x^2} + bx + c = 0$
We get, $a = 2,b = - 6,c = 3$
Thus, ${b^2} - 4ac = {( - 6)^2} - 4 \times 2 \times 3 = 36 - 24 = 12 > 0$
Thus the two roots are different real numbers.
Now we will apply Sri dharacharya’s formula.
The roots are given by,
$x = \dfrac{{ - ( - 6) \pm \sqrt {12} }}{{2 \times 2}}$
Split the roots,
$= \dfrac{{6 + \sqrt {12} }}{4},\;\dfrac{{6 - \sqrt {12} }}{4}$
Taking square root on both real root,
$= \dfrac{{6 + 2\sqrt 3 }}{4},\,\dfrac{{6 - 2\sqrt 3 }}{4}$
Taking common values and cancelling,
$= \dfrac{{3 + \sqrt 3 }}{2},{\text{ }}\dfrac{{3 - \sqrt 3 }}{2}$
Thus the real roots are $\dfrac{{3 + \sqrt 3 }}{2},\dfrac{{3 - \sqrt 3 }}{2}$.

Note: We have to concentrate on simplifying the roots to find the nature of the root of the equation. Because we may go wrong with the calculation of it.
Sridharacharya’s formula to solve quadratic equation
$a{x^2} + bx + c = 0$ is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The type of the roots of the quadratic equation is calculated by it’s discriminant ${b^2} - 4ac$.