Question

Describe quadratic formula with an example.

Answer 1

Dear Student,

Solution:

Quadratic formula is used to find the roots of a quadratic equation.

as we know that quadratic equation has two zeros.

we can find the value of zeros by putting the coefficients of different powers of x into the quadratic equation.

standard quadratic equation is given by equation
$a {x}^{2} + bx + c$
Quadratic formula : It is also known as sridharacharya formula.x 1,2

$= \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x1 = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \\ x2 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$
finding the nature of roots by determinant:

determinant is given as D
$= {b}^{2} - 4ac$
1) If D = 0, it means that equation has real and equal roots

2) If D > 0, it means that equation has real but distinct roots

3) If D <0, it represent that equation does not have any real.

Let
${x}^{2} + 4x + 3 = 0$
here a = 1

b= 4

c= 3

D = (4)^2 - 4(1)(3)

D = 16-12 =4 >0

that is equation has real and distinct roots,t these are
$x1 = \frac{ - 4 + \sqrt{4} }{2} \\ \\ = \frac{ - 4 + 2}{2} = - 1 \\ \\ x2 = \frac{ - 4 - \sqrt{4} }{2} \\ x2 = - 3$
Hope it helps you.

Answer 2

HELLO DEAR,

Quadratic formula is used to find the roots of a quadratic equation.

as we know that quadratic equation has two zeros.

general equation for using quadratic formula is;

ax² + bx + c = 0

NOW,

quadratic formula is;

$\sf{x = {-b{+\over} \sqrt{b^2 - 4ac}\over2a}}$

It is also known as sridharacharya rule

prove of sridharacharya rule:-

ax² + bx + c = 0

driving equation by "a"

x² + bx/a + c/a = 0

x² + bx/a + {(b/2a)² - (b/2a)²} + c/a = 0

(x + b/2a)² = b²/4a² - c/a

(x + b/2a)² = {b² - 4ac}/4a²

taking square root both side;

(x + b/2a) = ± $\sf{\sqrt{b^2 - 4ac}\over2a}$

taking (+ ve)

x = -b/2a + $\sf{\sqrt{b^2 - 4ac}\over2a}$

x = $\sf{{-b + \sqrt{b^2 - 4ac}\over2a}}$

taking (-ve)

x = -b/2a - $\sf{\sqrt{b^2 - 4ac}\over2a}$

x = $\sf{{-b - \sqrt{b^2 - 4ac}\over2a}}$

where, Determinants (D) = b² - 4ac

AND, b is the cofficient of x , a is the cofficient of x² , c is the constant or the cofficient of x^0

NOW,

finding the nature of roots by determinant:

determinant D

1) If D = 0, it means that equation has real and equal roots

2) If D > 0, it means that equation has real but distinct roots

3) If D <0, it represent that equation does not have any real.

e.g., x² + x - 2 = 0

where, a = 1 , b = 1 , c = -2,

so, D = b² - 4ac

D = (1)² - 4(1)(-2)

D = 1 + 8

D = 9

NOW,

using the foumula;

for (+ve)

x = $\sf{{-(1) + \sqrt{9}\over2*1}}$

x = $\sf{{-1 + 3\over2}}$

x = $\sf{2/2 = 1}$

for (-ve)

x = $\sf{{-(1) + \sqrt{9}\over2*1}}$

x = $\sf{{-1 - 3\over2}}$

x = -4/2 = -2

I HOPE ITS HELP YOU DEAR,

THANKS