Question

Find the condition that ax^3+bx^2+3cx+d=0 has its roots in AP

Answer 1

let the roots be x-y, x, x+y

$(x - y) + x + (x + y) = - \frac{b}{a} \\ 3x = - \frac{b}{a} \\x = - \frac{b}{3a} \\ x(x + y)(x - y) = - \frac{d}{a} \\ ( - \frac{b}{3a} )( {x}^{2} - {y}^{2} ) = - \frac{d}{a} \\ {x}^{2} - {y}^{2} = \frac{3d}{b} \\ {x}^{2} - {y}^{2} = \frac{3d}{b} \\ \frac{ {b}^{2} }{9 {a}^{2} } - {y}^{2} = \frac{3d}{b} \\ {y}^{2} = \frac{ {b}^{2} }{9 {a}^{2} } - \frac{3d}{b} \\ {y}^{2} = \frac{ {b}^{3} - 27 {a}^{2}d }{9 {a}^{2}b } \\ x(x + y) + x(x - y) + (x + y)(x - y) = \frac{c}{a} \\ 3 {x}^{2} - {y}^{2} = \frac{c}{a} \\ 3( \frac{ {b}^{2} }{9 {a}^{2} } ) - \frac{ {b}^{3} - 27 {a}^{2}d }{9 {a}^{2}b } = \frac{c}{a} \\ \frac{ {b}^{3} - {b}^{3} + 27 {a}^{2} d}{9 {a}^{2} b} = \frac{c}{a} \\ 3 \frac{d}{b} = \frac{c}{a} \\ \bold{\huge{\red{3ad - bc = 0}}}$

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