α10+β10=123α+β=1α9+β9=76α8+β8=47P10=P9+P8x2=x+1⇒x2−x−1=0α+β=1,αβ=−1α1+β1=αβα+β=−11=−1,αβ1=−1
Let P_n=α^n+β^n, n ∈ N. If P_10=123, P_9=76, P_8=47 and P_1=1, then the quadratic equation having roots 1 α and 1 β is :
JEE Main 2025 — Mathematics Algebra
Let Pn=αn+βn,n∈N. If P10=123,P9=76, P8=47 and P1=1, then the quadratic equation having roots α1 and β1 is :
Held on 2 Apr 2025 · Verified 6 Jul 2026.
Options
- A
x2−x+1=0
- B
x2+x−1=0
- C
x2−x−1=0
- D
x2+x+1=0
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