Question

Determine the nature of the roots of the
quadratic equation
${2 x }^{2} + 3x - 4 = 0$
​

Answer 1

Step-by-step explanation:

b^2-4ac=3^2-4×2×-4

=9+32

=41

SINCE discriminant is greater than zero therefore given quadratic equation will have two distinct real roots

Answer 2

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \bull \: \: \sf \: A \: quadratic \: equation \: {2x}^{2} + 3x - 4 = 0$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \bull \: \sf \: The \: nature \: of \: roots.$

$\large\underline{\sf{Solution-}}$

Concept Used :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

Let's solve the problem now!!

Given Quadratic equation is

$\: \: \: \: \: \bull \: \sf \: {2x}^{2} + 3x - 4 = 0$

On comparing with ax² + bx + c = 0, we get

• a = 2

• b = 3

• c = - 4

Now,

We know,

• Discriminant (D) of the quadratic equation is given by

$\rm :\longmapsto\:Discriminant \: (D) \: = \: {b}^{2} \: - \:4 ac$

On substituting the values of a, b and c, we get

$\rm :\longmapsto\:D = {(3)}^{2} - 4 \times 2 \times ( - 4)$

$\rm :\longmapsto\:D = 9 + 32$

$\rm :\longmapsto\:D = 41$

$\bf\implies \:Discriminant \: (D) > 0$

$\rm :\implies\:Roots \: of \: the \: equation \: are \: real \: and \: unequal.$