Question

 The difference between the roots of the equation 3x^2+bx+10=0 is equal to 4 1/3 . Find b.

Answer 1

`\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√(b^2-4(3)(10))}{2(3)}\qquad now \\ \quad \\ \cfrac{-b\pm√(b^2-4(3)(10))}{2(3)}\to \begin{cases} \cfrac{-b+√(b^2-4(3)(10))}{2(3)} \\ \quad \\ \cfrac{-b-√(b^2-4(3)(10))}{2(3)} \end{cases}\impliedby \textit{two roots}`


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

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