Answer:
a = 3
b = -7
c = 13
Step-by-step explanation:
[tex]\sf{given \ equation: 3x^2+bx+c[/tex]
[tex]\sf{sum \ of \ roots :\sf\frac{7}{3}[/tex]
[tex]\sf{product \ of \ the \ roots:\frac{13}{3}[/tex]
[tex]\sf{sum \ of \ roots \ = -\dfrac{coefficient \of \ x}{coefficient \ of \ x^2} }[/tex]
[tex]\sf \frac{-b}{3} = \frac{7}{3}[/tex]
[tex]\sf b = \frac{-7*3}{3}[/tex]
[tex]\sf b =- 7[/tex]
[tex]\sf{product \ of \ roots \ = \dfrac{constant \ term}{coefficient \ of \ x^2} }[/tex]
[tex]\sf{\frac{c}{3} =\frac{13}{3}[/tex]
[tex]\sf c = \frac{13*3}{3}[/tex]
[tex]\sf c = 13[/tex]
Therefore found that a = 3, b = -7, c = 13
Answer:
b = -7
c = 13
Step-by-step explanation:
[tex]3x^2+bx+c[/tex]
[tex]\textsf{sum of roots}=\dfrac{-b}{a}\\\\\textsf{product of roots}=\dfrac{c}{a}[/tex]
[tex]\textsf{Given sum of roots}=\dfrac73:[/tex]
[tex]\implies \dfrac{-b}{a}=\dfrac73[/tex]
[tex]\implies a=3, b=-7[/tex]
[tex]\textsf{Given product of roots}=\dfrac{13}{3}:[/tex]
[tex]\implies \dfrac{c}{a}=\dfrac{13}{3}[/tex]
[tex]\implies a=3, c=13[/tex]
Therefore,
[tex]3x^2-7x+13[/tex]
NB: The actual calculated roots are imaginary numbers as the function [tex]f(x)=3x^2-7x+13[/tex] does not intercept the x-axis
