Verified answer

Plug in x = 3 and y = 0:

0 = 45 + 3b + c

3b + c = -45

Find the zeros of the equation using the quadratic formula

x = (-b/10) +/- sqrt(b^2 - 20c)/10

x1 = (-b + sqrt(b^2 - 20c)/10)

x2 = (-b - sqrt(b^2 - 20c)/10)

x1 + x2 = 2(x1)(x2)

-2b/10 = 2(20c/100) = 2c/5

20c = -10b

10b + 20c = 0

Combine this with our previous equation above, we have two equations and two unknowns:

3b + c = -45

10b + 20c = 0

Divide equation 2 by 20

3b + c = -45

0.5b + c = 0

Subtract equatin 2 from equation 1

2.5b = -45

b = -45/2.5 = -18

c = 9

Check:

f(x) = 5x^2 - 18x + 9

f(x) = 5x^2 - 15x - 3x + 9

f(x) = 5x(x - 3) - 3(x - 3)

f(x) = (5x - 3)(x - 3)

Zeros are 3/5 and 3

Sum is 18/5, product is 9/5, so it works out.

f(3) = 0, then

45 + 3b + c = 0........[Eq1]

Let the roots be α, ß then we know that

α+ß = -b/5,

αß = c/5

Sum of zeroes is twice their product so

-b/5 = 2c/5........Eq2

Solve Eq1 and Eq2 simultaneously, then

b=-18, c=9