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Find the values of the coefficients b and c in the quadratic equation 3x2+bx+c if the sum and product of the roots of this equation are respectively 7/3 and 13/3.

User Stephen Rudolph
by
3.1k points

2 Answers

23 votes
23 votes

Answer:

b = -7

c = 13

Explanation:


3x^2+bx+c


\textsf{sum of roots}=(-b)/(a)\\\\\textsf{product of roots}=(c)/(a)


\textsf{Given sum of roots}=\frac73:


\implies (-b)/(a)=\frac73


\implies a=3, b=-7


\textsf{Given product of roots}=(13)/(3):


\implies (c)/(a)=(13)/(3)


\implies a=3, c=13

Therefore,


3x^2-7x+13

NB: The actual calculated roots are imaginary numbers as the function
f(x)=3x^2-7x+13 does not intercept the x-axis

User Caleb Nance
by
3.9k points
11 votes
11 votes

Answer:

a = 3

b = -7

c = 13

Explanation:


\sf{given \ equation: 3x^2+bx+c


\sf{sum \ of \ roots :\sf(7)/(3)


\sf{product \ of \ the \ roots:(13)/(3)

  • a = 3
  • b = b
  • c = c


\sf{sum \ of \ roots \ = -(coefficient \of \ x)/(coefficient \ of \ x^2) }


\sf (-b)/(3) = (7)/(3)


\sf b = (-7*3)/(3)


\sf b =- 7


\sf{product \ of \ roots \ = (constant \ term)/(coefficient \ of \ x^2) }


\sf{(c)/(3) =(13)/(3)


\sf c = (13*3)/(3)


\sf c = 13

Therefore found that a = 3, b = -7, c = 13

User Arnav
by
2.5k points
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