Question

If one root of the equation ax + bx +C =0 be the square of the other, prove that b' +ac+ací = 3abc.

JEE/Engineering Exams

Maths

Solution

212

1.0 (1 ratings)

64 one root of given equ ax ( ^{2}+b x+c=0 ). Be a and rinel be ( alpha^{2} ) -0 ( alpha+alpha^{2}=-frac{b}{a}-1 ) ( alpha cdot alpha^{2}=frac{c}{a} Rightarrow alpha^{3}=frac{c}{a}-1 ) Taking cube of 0 , ( (x+2)^{3}-4 ) an ( left(alpha+alpha^{2}right)^{3}=-frac{b^{3}}{a^{3}} ) ( alpha^{3}+alpha^{6}+3 alpha cdot alpha^{4}+3 alpha^{2} cdot alpha^{2}=frac{-b^{3}}{a^{3}} ) ( alpha^{3}+left(alpha^{3}right)^{2}+3 alpha^{3}left(alpha^{2}+alpharight)=-frac{b^{3}}{a^{3}} ) substituting ralue form 0 and (2) ( frac{c}{a}+left(frac{c}{a}right)^{2}+3 frac{c}{a}left(-frac{b}{a}right) cdot=-frac{b^{3}}{a^{3}} ) ( frac{c}{a}+frac{c^{2}}{a^{2}}-frac{3 b c}{a^{2}}=-frac{b^{3}}{a^{3}} ) multiplying equ. by a ( ^{3} ), we get ( a^{2} c+a c^{2}-3 a b c=-b^{3} ) ( Rightarrow quad b^{3}+a^{2} c+a c^{2}=3 a b c )