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senarillosk · Nov 15, 2015

Find the value of n for which the
equation (n + 2) x²+ 3x - 5 = 0 has one
root triple the other.

Would appreciate it if someone explains this to me. Thank you.

kawaiipotato · Nov 15, 2015

$(n + 2) x^ 2+ 3x - 5 = 0$

$$ has roots $: \alpha , \beta$

$But given that one root is three times the other, then$

$\alpha = 3 \beta$

$$roots$: \beta , 3 \beta$

$Using the relationships between coefficients of a polynomial and its roots,$

$$ Sum of roots $ = 4 \beta = \frac{-b}{a} = \frac{-3}{n+2}$

$$ Product of roots $ = 3 \beta ^2 = \frac{c}{a} = \frac{-5}{n+2}$

$Using the sum of roots, making $ \beta $ the subject and squaring and substituting into product of roots$

$3 * \frac{9}{16 (n+2)^2 } = \frac{-5}{n+2}$

$\frac{27}{-80} = n+2$

$n = \frac{-27}{80} - 2 = \frac{-187}{80}$

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senarillosk

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