Verified Solution • 2025–26 Session moderate

QJEEM23Mae21f: JEE Main 2023 Mathematics - Class 11 JEEM23Mae21f Solution

If aa and bb are the roots of the equation x2−7x−1=0x^{2}-7 x-1=0 , then the value of a21+b21+a17+b17a19+b19\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}} is equal to _____________.

Step-by-Step Verified Solution

We have, aa and bb are the roots of the equation

amp;x2−7x−1=0amp;⇒a2−7a−1=0⇒a2−1=7a.........(i)\begin{aligned}& x^2-7 x-1=0 \\\\& \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i)\end{aligned}

On squaring both sides, we get a4+1=51a2a^4+1=51 a^2

Similarly, b4+1=51b2b^4+1=51 b^2 ...........(ii)

 Now, a21+b21+a17+b17a19+b19=a17(a4+1)+b17(b4+1)a19+b19\text { Now, } \frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}}

amp;=a17(51a2)+b17(51b2)a19+b19[∵ From Eq. (i) and (ii) ]amp;=51[a19+b19]a19+b19=51\begin{aligned}& =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\\\& =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring