Question

Find the nature of the roots of the following quadratic equation. If the real roots exist, find them.
$2{x^2} - 3x + 5 = 0$
A) x = 0 and x = -2
B) x = 3, x = -6
C) No real root
D) None of these

Answer 1

Hint: The nature of the roots depends on the value of the discriminant of the quadratic equation.
$a{x^2} + bx + c = 0$, where $a \ne 0$
Find the Discriminant, $D = {b^2} - 4ac$ , of the given quadratic equation, and check the sign (i.e. positive or negative or zero) to know if there are two solutions or one solution or no solution.

Complete step-by-step answer:
Step 1: Given the quadratic equation:
$2{x^2} - 3x + 5 = 0$
On comparing with standard quadratic equation: $a{x^2} + bx + c = 0$, where $a \ne 0$
a = 2, b = -3, c = 5
Step 2: Find discriminant:
$D = {b^2} - 4ac$
$D = {\left( { - 3} \right)^2} - 4 \times 2 \times 5$
$
   \Rightarrow {\text{ }} = 9 - 40 \\
   \Rightarrow {\text{ }} = - 31 \\
 $
Step 3: Check the sign of discriminant:
$D < 0$
Hence, the roots are imaginary.
Final answer: The roots of $2{x^2} - 3x + 5 = 0$ are not real. Thus the correct option is (C).

Additional Information:
Roots of the quadratic equation is given by:
Quadratic equation: $a{x^2} + bx + c = 0$, where $a \ne 0$
Roots: \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
The imaginary roots of the given quadratic equation are:
$2{x^2} - 3x + 5 = 0$
D = -31
\[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
$
  {\text{ }}x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {\left( { - 31} \right)} }}{{2\left( 2 \right)}} \\
   \Rightarrow {\text{ }} = \dfrac{{3 \pm {\text{i}}\sqrt {31} }}{4} \\
 $

Note: For quadratic equation: $a{x^2} + bx + c = 0$, where $a \ne 0$
Let $y = f\left( x \right) = a{x^2} + bx + c = 0$
Discriminant, $D = {b^2} - 4ac$
A discriminant of zero indicates that the quadratic has a repeated real number solution.
i.e. $D = 0$ , roots are real and equal.
$ \Rightarrow {b^2} - 4ac = 0$

A positive discriminant indicates that the quadratic has two distinct real number solutions.
i.e. $D > 0$ , roots are real and unequal.
$ \Rightarrow {b^2} - 4ac > 0$

A negative discriminant indicates that neither of the solutions is real numbers.
And if D < 0, as in the case of the given question, roots are imaginary.
$ \Rightarrow {b^2} - 4ac < 0$