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# 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 = -2B) x = 3, x = -6C) No real rootD) None of these

Last updated date: 22nd Jun 2024
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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.

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).

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.
$\Rightarrow {b^2} - 4ac < 0$