Question

write an algorithm to evaluate following expression ax2+bx+c=0​

Answer 1

Answer:

#include <stdio.h>

#include <math.h>

int main() {

float a, b, c, discriminant, x1, x2, r, i;

printf("coefficient of x^2: ");

scanf("%f", &a);

printf("coefficient of x: ");

scanf("%f", &b);

printf("constant term: ");

scanf("%f", &c);

discriminant = pow(b,2) - 4*a*c;

if(discriminant > 0) {

 x1 = (-b + sqrt(discriminant))/(2*a);

 x2 = (-b - sqrt(discriminant))/(2*a);

 printf("x1 = %.2f \n", x1);

 printf("x2 = %.2f \n", x2);

} else if (discriminant == 0) {

 x1 = -b/(2*a);

 x2 = -b/(2*a);

 printf("x1 = %.2f \n", x1);

 printf("x2 = %.2f \n", x2);

} else {

 r = -b/(2*a);

 i = sqrt(-discriminant)/(2*a);

 printf("x1 = %.2f +i %.2f \n", r, i);

 printf("x2 = %.2f -i %.2f \n", r, i);

}

return 0;

}

Explanation:

Answer 2

ax2 + bx + c

where,

a, b, and c are coefficient and real numbers and also a ≠ 0.

If a is equal to 0 that equation is not valid quadratic equation.

we know , if n² = a

then , n = ±√a

We will try ton comert ax²+ba+c=0, a ≠ 0 into something like x²

Divide both sides by a

${x}^{2} x \times \frac{b}{a} + \frac{c}{a} = 0$

${x}^{2} + x \times \frac{b}{a} = - \frac{c}{a}$

$(x\ + \frac{b}{2a} )^{2} - \frac{b}{4a} ^{2} = - \frac{c}{a}$

$(x \times \frac{b}2a)^{2} = \frac{ {b}^{2} - 4ac}{4a}$

Now this equation gives two roots

$n = \frac{ - 6 \sqrt{6 ^{2} - 4ac} }{2a}$

or

$n = \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

#SPJ2