Question

under what condition of the roots of the equation ax²+by+c=0 are such that one root will be the square of other​

Answer 1

Step-by-step explanation:

$let \: one \: root \: be \: \alpha \\ \:and \: the \: other \: root \: will \: be \: n \alpha \: \\ by \: the \: given \: condition \\ \\ sum = \\ \: \alpha + n \alpha = \frac{ - b}{a} \: \:( or) \: \: \alpha = \frac{b}{a(1 + n)} \: \: \: \: eq1 \\ \\ Product = \\ n { \alpha }^{2} = \frac{c}{a} ==> \: \alpha = \frac{c}{na} \: \: eq2 \\ \\ Now \\ \\ from \: eq \: 1 \: and \: eq \: 2 \\ \\ { \alpha }^{2} = ( { \alpha }^{2} ) \\ \\ ==> \: \frac{c}{an} = \frac{ {b}^{2} }{ {a}^{2} {(1 + n)}^{2} } \\ \\ ==> n{b}^{2} = ac {(n + 1)}^{2}$