An algebra problem by odyson santos

let $x_1$ and $x_2$ be the roots of the quadratic equation $ax^2+bx+c=0$ . By Vieta's Formula , the sum of the roots is $x_1+x_2=-\dfrac{b}{a}$ and the product of the roots is $x_1(x_2)=\dfrac{c}{a}$ . But the given in the problem is the sum of the reciprocals of the roots, the sum of the reciprocals of the roots is

$\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1(x_2)}=\dfrac{sum~of~the~roots}{product~of~the~roots}$

The sum of the roots is $x_1+x_2=-\dfrac{7}{3}$ while the product of the roots is $x_1(x_2)=\dfrac{k}{3}$

Substituting, we have

$\dfrac{7}{3}=\dfrac{-\dfrac{7}{3}}{\dfrac{k}{3}}=-\dfrac{7}{3}\left(\dfrac{3}{k}\right)$ $\implies$ $\dfrac{7}{3}=-\dfrac{7}{k}$ $\implies$ $\dfrac{1}{3}=\dfrac{1}{k}$ $\implies$ $\boxed{k=-3}$

a=3,b=7,c=k

$Sum of roots= \frac{-b}{a}=\frac{-7}{3}$

$Product of roots=\frac{c}{a}=\frac{k}{3}$

Let the roots of the equations be α and β

Sum of reciprocal of the roots of the equations

$=\frac{1}{α}+\frac{1}{β}$

$=\frac{α+β}{αβ}$

$=\frac{-7}{3}/ \frac{k}{3}$

$=\frac{-7}{3}\times(\frac{3}{k})$

$\frac{-7}{3}\times \frac{3}{k}=\frac{7}{3}$

$\frac{-63}{9k}=\frac{21k}{9k}$

$-63=21k$

$k=-3$

3 (1/ y)^2 + 7 (1/ y) + k = 0

k y^2 + 7 y + 3 = 0

y^2 + (7/ k) y + 3/ k = 0

Sum of reciprocal = -7/ k = 7/ 3

k = -3

We can check by substitution that 3 x^2 + 7 x - 3 = 0 do give a sum of reciprocal equals to 7/ 3.

1/a +1/b =7/3 Thus , (a+b)/ab =7/3 but we know that a+b =-7/3 Therefore we conclude that ab=-1 Solving for k k/3 =-1 ( Product of roots criterion) k=-3

By Vieta's theorem: k = -ab but 1/a + 1/b = 7/3 since ab = -3 thus k = -3