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:
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Step-by-Step Solution
Step 1: Establish the recurrence relation
We are given values for Pn for different n. This suggests that α and β are roots of a quadratic equation. Let the quadratic equation be x2−Ax−B=0. Then, by Vieta's formulas, α+β=A and αβ=−B. The sequence Pn=αn+βn satisfies the recurrence relation Pn=APn−1+BPn−2. We can use the given values to find A and B.
Step 2: Formulate system of equations
Using the given values P10=123, P9=76, and P8=47, we can substitute them into the recurrence relation to form a system of linear equations. For n=10, we have P10=AP9+BP8, which gives 123=76A+47B. For n=9, we have P9=AP8+BP7. However, we don't have P7. Let's use P9=AP8+BP7 and P8=AP7+BP6. We need to find P7 and P6 or use a different approach. Let's re-evaluate the system using the given values directly.
Step 3: Solve for A and B
We are given P1=1. Since P1=α1+β1=α+β, and we know that for the quadratic equation x2−Ax−B=0, the sum of roots is A=α+β, we can directly deduce that A=1. Now, substitute A=1 into the first equation 123=76A+47B. This gives 123=76(1)+47B, which simplifies to 47=47B, so B=1.
Step 4: Form the original quadratic equation
With A=1 and B=1, the quadratic equation whose roots are α and β is x2−Ax−B=0, which becomes x2−1x−1=0, or simply x2−x−1=0.
Step 5: Find the quadratic equation with roots α1 and β1
If a quadratic equation has roots α and β, then the quadratic equation with roots α1 and β1 can be found by replacing x with x1 in the original equation. So, substitute x=y1 into x2−x−1=0. This gives (y1)2−(y1)−1=0. Multiplying by y2 (assuming y=0), we get 1−y−y2=0, which can be rewritten as y2+y−1=0.