Quadratic Equations – Recurrence Relations – JEE Main 2025 Shift 2

Solution:

Given the quadratic equation $x^2 – ax – b = 0$, its roots $\alpha$ and $\beta$ satisfy the recurrence relation (Newton’s Sums):
$$P_n = a P_{n-1} + b P_{n-2}$$

We are given the values for $n=5$ and $n=6$.
Using $n=5$:
$$P_5 = a P_4 + b P_3$$
$$11\sqrt{7}i = a(-3\sqrt{7}i) + b(-5\sqrt{7}i)$$
Dividing by $\sqrt{7}i$:
$$11 = -3a – 5b \quad \dots(1)$$

Using $n=6$:
$$P_6 = a P_5 + b P_4$$
$$45\sqrt{7}i = a(11\sqrt{7}i) + b(-3\sqrt{7}i)$$
Dividing by $\sqrt{7}i$:
$$45 = 11a – 3b \quad \dots(2)$$

Now we solve the system of linear equations for $a$ and $b$:
1. $3a + 5b = -11$
2. $11a – 3b = 45$

Multiply equation (1) by 3 and equation (2) by 5:
$$9a + 15b = -33$$
$$55a – 15b = 225$$
Adding them:
$$64a = 192 \Rightarrow a = 3$$
Substitute $a=3$ into equation (1):
$$3(3) + 5b = -11 \Rightarrow 9 + 5b = -11 \Rightarrow 5b = -20 \Rightarrow b = -4$$

From the quadratic equation $x^2 – ax – b = 0$, the sum and product of roots are:
$$\alpha + \beta = a = 3$$
$$\alpha\beta = -b = -(-4) = 4$$

We need to find $|\alpha^4 + \beta^4|$. First, find $\alpha^2 + \beta^2$:
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = (3)^2 – 2(4) = 9 – 8 = 1$$

Now find $\alpha^4 + \beta^4$:
$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 – 2(\alpha\beta)^2$$
$$\alpha^4 + \beta^4 = (1)^2 – 2(4)^2 = 1 – 32 = -31$$

The absolute value is:
$$|\alpha^4 + \beta^4| = |-31| = 31$$

Ans. (31)