Equation x2 plus Bx plus c equals 0 has 5 as the sum of its roots and 15 as the sum of the squares of its roots The value of 'c' is equals?

Answer 1

x2 - (-b/a)x + (c/a) = 0 or

x2 - (sum of the roots)x + (product of the roots) = 0

Let the roots be r1 and r2. So we have:

r1 + r2 = 5

(r1)2 + (r2)2 = 15

r1 = 5 - r2 (express r1 in term of r2)

(5 - r2)2 + (r2)2 = 15

25 - 10r2 + (r2)2 + (r2)2 = 15

2(r2)2 - 10r + 25 = 15 (subtract 15 to both sides)

2(r2)2 - 10r + 10 = 0 (divide by 2 to both sides)

(r2)2 - 5r + 5 = 0 (use the quadratic formula)

r2 = [-b + &- sq root of (b - 4ac)]/2a

r2 = {-(-5) + &- sq root of [(-5)2 - 4(1)(5)]}/2(1) = [5 + &- sq root of (25 - 20)]/2 = (5 + &- sq root of 5)/2

r1 = 5 - r2

r1 = 5 - (5 + &- sq root of 5)/2

Thus, when r2 = (5 + sq.root of 5)/2, r1 = (5 - sq.root of 5)/2 or vice versa.

Since the given equation is x2 + bx + c = 0, a = 1, then c equals to the product of roots.

So that,

c = (r1)(r2) = [(5 - sq.root of 5)/2][(5 + sq.root of 5)/2] = [52 - (sq.root of 5)2]/4 = 5