Question

Find the nature of roots of the following quat quadratin equation : 2x² 6x + 3 =0​

Answer 1

Quadratic Equations

A quadratic equation in a variable $x$ is an equation which is of the form $ax^2 + bx + c = 0$ where constants $a$, $b$ and $c$ are all real numbers and $a \neq 0$.

In case of a quadratic equation $ax^2 + bx + c = 0$ the expression $b^2 - 4ac$ is called the discriminant.

Let's head to the question now.

The given expression is;

$\qquad 2x^2 + 6x + 3 = 0$

Now, comparing the given equation with the standard form of quadratic equation, we get:

$\qquad a = 2, \: b = 6, \: c = 3$

Now using the discriminant formula and solving the equation, we get:

$\implies 6^2 - 4 \times 2 \times 3 \\ \\ \implies 36 - 4 \times 2 \times 3 \\ \\ \implies 36 - 24 \\ \\ \implies 12$

As we know that, If Discriminant, $D > 0$, then roots of the equation are real and unequal.

Hence, the nature of the roots of the quadratic equation are real and unequal.

Answer 2

$\sf \ \underline { \red{solution:-}}$

Step:-1

• $\sf \: compare \: with \rightarrow \: a {x }^{2} + bx + c = 0 \\ \sf \: a = 2 \: and \: b = - 3 \: and \: c = 5$

Step:-2

• $\sf calculate \rightarrow \: {b}^{2} - 4ac \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 - 40 = - 3 1$

Hence,

• Roots are imaginary.