Given that the difference between the roots of the equation [tex]3x^2+bx+10=0[/tex] is [tex]4 \frac{1}{3} = \frac{13}{3} [/tex].
Recall that the sum of roots of a quadratic equation is given by [tex]- \frac{b}{a} [/tex].
Let the two roots of the equation be [tex] \alpha [/tex] and [tex] \alpha + \frac{13}{3} [/tex], then
[tex] \alpha + \alpha + \frac{13}{3} =2 \alpha + \frac{13}{3} =- \frac{b}{a} =- \frac{b}{3} \\ \\ i.e.\ \ 2 \alpha + \frac{13}{3}=- \frac{b}{3}[/tex] . . . (1)
Also recall that the product of the two roots of a quadratic equation is given by [tex] \frac{c}{a} [/tex], thus:
[tex] \alpha \left( \alpha + \frac{13}{3} \right)= \alpha ^2+ \frac{13}{3} \alpha = \frac{c}{a} = \frac{10}{3} \\ \\ i.e.\ \ \alpha ^2+ \frac{13}{3} \alpha=\frac{10}{3}[/tex] . . . (2)
From (1), we have:
[tex]2 \alpha =- \frac{b}{3} - \frac{13}{3} \\ \\ \Rightarrow \alpha =- \frac{b}{6} - \frac{13}{6} [/tex]
Substituting for alpha into (2), gives:
[tex]\left(- \frac{b}{6} - \frac{13}{6}\right)^2+ \frac{13}{3} \left(- \frac{b}{6} - \frac{13}{6}\right)= \frac{10}{3} \\ \\ \Rightarrow \frac{b^2}{36} + \frac{13b}{18} + \frac{169}{36} - \frac{13b}{18} - \frac{169}{18} = \frac{10}{3} \\ \\ \Rightarrow \frac{b^2}{36} - \frac{289}{36} =0 \\ \\ \Rightarrow \frac{b^2}{36} = \frac{289}{36} \\ \\ \Rightarrow b^2=289 \\ \\ \Rightarrow b=\pm\sqrt{289}=\pm17[/tex]
