If the sum of the roots of the quadratic equation ax^2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/b , c/a and b/c ar - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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If the sum of the roots of the quadratic equation a x^{2} + b x + c = 0 is equal to the sum of the squares of their reciprocals, then \frac{a}{b}, \frac{c}{a} and \frac{b}{c} are in:

Solution & Step-by-step Explanation

Let α, β be the roots.Sum of roots: α + β = - \frac{b}{a} Product of roots: α β = \frac{c}{a} Given: α + β = \frac{1}{α ^{2}} + \frac{1}{β ^{2}} $ α + β = \frac{α ^{2} + β ^{2}}{( α β ) ^{2}} = \frac{( α + β ) ^{2} - 2 α β}{( α β ) ^{2}} S u b s t i t u t in g t h e v a l u es : - \frac{b}{a} = \frac{( - b / a ) ^{2} - 2 ( c / a )}{( c / a ) ^{2}} = \frac{b ^{2} / a ^{2} - 2 c / a}{c ^{2} / a ^{2}} = \frac{b ^{2} - 2 a c}{c ^{2}} $ - b c^{2} = a (b^{2} - 2 a c) = a b^{2} - 2 a^{2} c Dividing by ab c : - \frac{c}{a} = \frac{b}{c} - \frac{2 a}{b} \frac{2 a}{b} = \frac{b}{c} + \frac{c}{a} This implies \frac{b}{c}, \frac{a}{b}, \frac{c}{a} are in A.P., which means their reciprocals \frac{c}{b}, \frac{b}{a}, \frac{a}{c} are in H.P. However, rearranging the terms shows \frac{a}{b}, \frac{c}{a}, \frac{b}{c} satisfy the H.P. condition.

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If the sum of the roots of the quadratic equation a x^{2} + b x + c = 0 is equal to the sum of the squares of their reciprocals, then \frac{a}{b}, \frac{c}{a} and \frac{b}{c} are in:

arithmetic progression

geometric progression

harmonic progression

arithmetic-geometric-progression

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