InterviewSolution
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InterviewSolution
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d, e, and f are in GP |
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Answer» d, e and f are in GP `:. ""b^(2) = ac`..(i) and the given quadratic equations `AX^(2) + 2b x + c = 0`...(i) `dx^(2) + 2ex + f = 0`...(III) For quadratic Eq. (ii) , the DISCRIMINANT `D = (2b)^(2) - 4ac` `= 4 (b^(2) - ac) = 0` [from Eq. (i)] `rArr` Quadratic Eq. (ii) have equal roots, and it is equal to `x = -(b)/(a)`, and it is given that quadratic Eqs. (ii) and (iii) have a common root, so `d (-(b)/(a))^(2) + 2e (-(b)/(a)) + f = 0` `rArr db^(2) - 2eba + a^(2) f = 0` `rArr d (ac) - 2eab + a^(2) f = 0 "" [ :' b^(2) = ac]` `rArr d (ac) - 2eab + a^(2) f = 0 "" [ :' a != 0]` `rArr 2eb = dc + af` `rArr 2 (e)/(b) = (dc)/(b^(2)) + (af)/(b^(2))` [dividing each term by `b^(2)`] `rArr 2 ((e)/(b)) = (d)/(a) + (f)/(c) "" [ :'b^(2) = ac]` So, `(d)/(a), (e)/(b), (f)/(c)` are in AP. Alternate Solution Given, three distinct numbers a, b and c are in GP. Let `a = a, b = ar, c = ar^(2)` are in GP, which satisfies `ax^(2) + 2bx + c = 0` `:. ax^(2) + 2(ar) x + ar^(2) = 0` `rArr x^(2) + 2rx + r^(2) = 0 "" [ :' a != 0]` `rArr (x + r)^(2) = 0 rArr x = -r` According to the equation, `ax^(2) + 2bx + c = 0 and dx^(2) + 2ex + f = 0` have a common root. So, `x = -r` satisfies `dx^(2) + 2ex + f = 0` `:. d(-r)^(2) + 2e(-r) + f = 0` `rArr dr^(2) - 2er + f = 0` `rArr d((c)/(a)) - 2e ((c)/(b)) + f = 0` `rArr (d)/(a) - (2e)/(b) + (f)/(c) = 0` `rArr (d)/(a) + (f)/(c) = (2e)/(b) "" [ :' c != 0]` |
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