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Dafna1 [17]
4 years ago
11

The difference between the roots of the equation 3x2+bx+10=0 is equal to 4 1 3 . find b.

Mathematics
1 answer:
inna [77]4 years ago
5 0
Given that the difference between the roots of the equation 3x^2+bx+10=0 is 4 \frac{1}{3} = \frac{13}{3}.

Recall that the sum of roots of a quadratic equation is given by - \frac{b}{a}.

Let the two roots of the equation be \alpha and \alpha + \frac{13}{3}, then 

\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} . . . (1)

Also recall that the product of the two roots of a quadratic equation is given by \frac{c}{a}, thus:

\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} . . . (2)

From (1), we have:

2 \alpha =- \frac{b}{3} - \frac{13}{3}  \\  \\ \Rightarrow \alpha =- \frac{b}{6} - \frac{13}{6}

Substituting for alpha into (2), gives:

\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
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