JEE Mathematics 2025 Question with Solution

Given: $\alpha, \beta$ are roots of $x^2 - ax - b = 0$ and $P_n = \alpha^n - \beta^n$.

Given values are:

• $P_3 = -5\sqrt{7}$
• $P_4 = -3\sqrt{7}$
• $P_5 = 11\sqrt{7}$
• $P_6 = 45\sqrt{7}$

Find: $|\alpha^4 + \beta^4|$

Using properties of roots for $x^2 - ax - b = 0$:

• $\alpha + \beta = a$
• $\alpha\beta = b$

Also, the recurrence relation is

$$P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2}$$

so

$$P_n = aP_{n-1} - bP_{n-2}$$

From $P_5 = aP_4 - bP_3$,

$$11\sqrt{7} = a(-3\sqrt{7}) - b(-5\sqrt{7})$$ $$11 = -3a + 5b$$

From $P_6 = aP_5 - bP_4$,

$$45\sqrt{7} = a(11\sqrt{7}) - b(-3\sqrt{7})$$ $$45 = 11a + 3b$$

So we get the system

$$-3a + 5b = 11$$ $$11a + 3b = 45$$

Multiply the first equation by $3$ and the second by $5$:

$$-9a + 15b = 33$$ $$55a + 15b = 225$$

Subtracting gives

$$64a = 192$$ $$a = 3$$

Substitute in $-3a + 5b = 11$:

$$-3(3) + 5b = 11$$ $$-9 + 5b = 11$$ $$5b = 20$$ $$b = 4$$

Now,

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = a^2 - 2b = 3^2 - 2\cdot 4 = 1$$

Then

$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$$ $$\alpha^4 + \beta^4 = 1^2 - 2\cdot 4^2 = 1 - 32 = -31$$

Therefore,

$$|\alpha^4 + \beta^4| = 31$$

So the required numerical value is $31$.

Given: $\alpha, \beta$ are roots of $x^2 - ax - b = 0$ and $P_n = \alpha^n - \beta^n$.

Find: $|\alpha^4 + \beta^4|$

For roots of the quadratic equation,

$$\alpha + \beta = a, \qquad \alpha\beta = b$$

The sequence satisfies

$$P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2}$$

Using the given terms:

$$P_5 = aP_4 - bP_3$$ $$11\sqrt{7} = a(-3\sqrt{7}) - b(-5\sqrt{7})$$ $$11 = -3a + 5b$$

Similarly,

$$P_6 = aP_5 - bP_4$$ $$45\sqrt{7} = a(11\sqrt{7}) - b(-3\sqrt{7})$$ $$45 = 11a + 3b$$

Solve:

$$-3a + 5b = 11$$ $$11a + 3b = 45$$

After elimination,

$$a = 3, \qquad b = 4$$

Now compute

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 9 - 8 = 1$$

Hence,

$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2$$ $$\alpha^4 + \beta^4 = 1^2 - 2(4)^2 = -31$$

Therefore, the required value is $31$.

Note: The alternative solution text on the page states $\alpha\beta = -b$, which is inconsistent with the polynomial $x^2 - ax - b = 0$. The correct relation used above is $\alpha\beta = b$, and it matches the final answer.