Question

Solve the Equation :

x²+ 3x + 5 = 0 and find it's roots ​

Answer 1

Answer:

a =1 b =3 and c =5

b2-4ac

(3)2-4×1×5

9-20

-11 ans

Answer 2

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

$\rm :\longmapsto\: {x}^{2} + 3x + 5 = 0$

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

$\: \: \: \: \: \: \: \: \: \: \bull \: \sf \: \: The \: value \: of \: x$

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

The given Quadratic Equation,

$\rm :\longmapsto\: {x}^{2} + 3x + 5 = 0$

$\rm :\longmapsto\:On \: comparing \: with \: {ax}^{2} + bx + c = 0 \: \: we \: get$

$\: \: \: \: \: \: \: \: \: \: \bull \: \sf \: \: a = 1$

$\: \: \: \: \: \: \: \: \: \: \bull \: \sf \: \: b = 3$

$\: \: \: \: \: \: \: \: \: \: \bull \: \sf \: \: c = 5$

We have to first find Discriminant (D) of above equation.

So, Discriminant is given by

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

$\rm :\longmapsto\:Discriminant (D) = {(3)}^{2} - 4 \times 5 \times 1$

$\rm :\longmapsto\:Discriminant (D) = 9 - 20$

$\rm :\longmapsto\:Discriminant (D) = - 11$

$\bf\implies \:The \: {eq}^{n} \: has \: non - real \: roots \: given \: by$

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{D} }{2a}$

$\rm :\longmapsto\:x \: = \: \dfrac{ - 3 \: \: \pm \: \sqrt{ - 11} }{2 \times 1}$

$\rm :\longmapsto\:x \: = \: \dfrac{ - 3 \: \: \pm \: i \: \sqrt{ 11} }{2}$

Additional Information :-

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