Gujarati
8. Sequences and Series
hard

यदि त्रिघातीय समीकरण $a{x^3} + b{x^2} + cx + d = 0$ के मूल गुणोत्तर श्रेणी में हैं, तब

A

${c^3}a = {b^3}d$

B

$c{a^3} = b{d^3}$

C

${a^3}b = {c^3}d$

D

$a{b^3} = c{d^3}$

Solution

(a) माना समीकरण $a{x^3} + b{x^2} + cx + d = 0$ के मूल $\frac{A}{R},\;A,\;AR$ है

तब ${A^3} = $ मूलों का गुणनफल $ =  – \frac{d}{a}$

$ \Rightarrow $ $A =  – {\left( {\frac{d}{a}} \right)^{1/3}}$

समीकरण का एक मूल $A$ है

$\therefore a{A^3} + b{A^2} + cA + d = 0$

$ \Rightarrow $ $a\left( { – \frac{d}{a}} \right) + b{\left( { – \frac{d}{a}} \right)^{2/3}} + c{\left( { – \frac{d}{a}} \right)^{1/3}} + d = 0$

$b{\left( {\frac{d}{a}} \right)^{2/3}} = c{\left( {\frac{d}{a}} \right)^{1/3}}$ Þ ${b^3}\frac{{{d^2}}}{{{a^2}}} = {c^3}\frac{d}{a}$

${b^3}d = {c^3}a$.

Standard 11
Mathematics

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