An algebra problem by A Former Brilliant Member

In quadratic eqn. $ax^2+bx +c=0$ , the product of the roots is $\frac{c}{a}$ . In this case the product is $-\frac52$ . The product of the reciprocals of the roots will be $\frac{a}{c}=-\frac25$ . Only the third eqn. satisfies this. Note that if the sum of the roots is $m+n=-\frac{b}{a}$ , then the sum of the reciprocal roots will be $\frac{1}{m}+\frac{1}{n}=\frac{m+n}{mn}=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c}$ . In this case it would be $-\frac35$ , which matches the $-\frac{b}{a}$ in the third eqn. One can find the equation directly by using $x^2+\frac{b}{c}x+\frac{a}{c}=0$ . Multiply by $c$ to simplify to: $cx^2+bx+a=0$ or $-5x^2-3x+2=0$ or $\boxed{5x^2+3x-2=0}$

The roots of the original equation are $-1$ and $5/2$

The reciprocals of these are $-1$ and $2/5$ , so your equation is $(x + 1)(5x-2) = 0$

Multiplying out, you get $5x^2 + 3x - 2 = 0$