Question Number 202297 by hardmath last updated on 24/Dec/23
$$\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{a}\:=\:\mathrm{5} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{1}−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{2}−\boldsymbol{\mathrm{b}}} }{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} }\:=\:? \\ $$
Answered by MATHEMATICSAM last updated on 24/Dec/23
![((a^(−b ) + a^(1 − b) + a^(2 − b) )/a^(−b) ) = ((a^(−b) (1 + a + a^2 ))/a^(− b) ) = a^2 + a + 1 = 5 + 1 [∵ a^2 + a = 5] = 6](https://www.tinkutara.com/question/Q202301.png)
$$\frac{{a}^{−{b}\:} +\:{a}^{\mathrm{1}\:−\:{b}} \:+\:{a}^{\mathrm{2}\:−\:{b}} }{{a}^{−{b}} } \\ $$$$=\:\frac{\cancel{{a}^{−{b}} }\left(\mathrm{1}\:+\:{a}\:+\:{a}^{\mathrm{2}} \right)}{\cancel{{a}^{−\:{b}} }} \\ $$$$=\:{a}^{\mathrm{2}} \:+\:{a}\:+\:\mathrm{1} \\ $$$$=\:\mathrm{5}\:+\:\mathrm{1}\:\left[\because\:{a}^{\mathrm{2}} \:+\:{a}\:=\:\mathrm{5}\right] \\ $$$$=\:\mathrm{6} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Dec/23
$$\:\:\frac{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{1}−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{2}−\boldsymbol{\mathrm{b}}} }{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} }\: \\ $$$$=\mathrm{1}+{a}+{a}^{\mathrm{2}} =\mathrm{1}+\mathrm{5}=\mathrm{6} \\ $$