Respuesta :

[tex]\textit{quadratic formula}\\ {{ 3}}x^2{{ +b}}x{{ +10}}=0 \qquad \qquad x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}} \\ \quad \\ meaning\implies x=\cfrac{-b\pm\sqrt{b^2-4(3)(10)}}{2(3)}\qquad now \\ \quad \\ \cfrac{-b\pm\sqrt{b^2-4(3)(10)}}{2(3)}\to \begin{cases} \cfrac{-b+\sqrt{b^2-4(3)(10)}}{2(3)} \\ \quad \\ \cfrac{-b-\sqrt{b^2-4(3)(10)}}{2(3)} \end{cases}\impliedby \textit{two roots}[/tex]

[tex]\textit{and their root difference is }4\frac{1}{3}\implies \cfrac{13}{3}\qquad so \\ \quad \\ \left[ \cfrac{-b+\sqrt{b^2-4(3)(10)}}{2(3)} \right]-\left[ \cfrac{-b-\sqrt{b^2-4(3)(10)}}{2(3)} \right]=\cfrac{13}{3} \\ \quad \\ \left[ \cfrac{-b+\sqrt{b^2-4(3)(10)}}{6} \right]+\left[ \cfrac{+b+\sqrt{b^2-4(3)(10)}}{6} \right]=\cfrac{13}{3} \\ \quad \\ \cfrac{-b+\sqrt{b^2-4(3)(10)}+b+\sqrt{b^2-4(3)(10)}}{6}=\cfrac{13}{3} \\ \quad \\ \cfrac{2\sqrt{b^2-4(3)(10)}}{6}=\cfrac{13}{3}[/tex]

and I'm pretty sure you can take it from there

Otras preguntas