practice problems from PYQs.

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Q.
$x^4 - qx^3 + 2(10-a)x - 9qx + a = 0$

For $x$ where $a$ is real parameter,
has no real roots for $a \in (-\infty, -\tfrac{p}{q})$.

$p, q$ are co-prime. Find $p+q$.

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Solution:
Divide by x^2:

$$\frac{x^4 - 9x^3 + 2(10 - a)x^2 + 9ax + a^2}{x^2} = 0$$

$$x^2 - 9x + 20 - 2a + \frac{9a}{x} + \frac{a^2}{x^2} = 0$$

$$\left(x^2 + \frac{a^2}{x^2}\right) - 9\left(x - \frac{a}{x}\right) + 20 - 2a = 0$$

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Let x - a/x = k

$$\Rightarrow \quad x^2 + \frac{a^2}{x^2} - \frac{2a}{x} \cdot x = k^2$$

Simplifies to:

$$x^2 + \frac{a^2}{x^2} = k^2 + 2a$$

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Simplified and Substituted Expression:

$$k^2 + 2a - 9k + 20 - 2a = 0$$

Cancelling $+2a$ and $-2a$:

$$k^2 - 9k + 20 = 0$$

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Factored Form:

$$(k - 4)(k - 5) = 0 \quad \Rightarrow \quad k = 4 \quad \text{or} \quad k = 5$$

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Case 1: k = 4

x - a/x = 4 => x^2 - a = 4x => x^2 - 4x - a = 0

To ensure no real solutions:

$$D < 0$$

Discriminant of quadratic $x^2 - 4x - a = 0$ is:

$$D = 16 + 4a < 0$$

Solve:

$$4a < -16 \Rightarrow a < -4$$

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Final Result:

$$\boxed{a < -4}$$

Case 2:
k = 5
x - a/x = 5 => x^2 - a - 5x = 0  D< 0 => a < -25/4
Combining both cases, finally:
a < -25/4
So p = 25, q = 4

Q:

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Given:

• $a, b \in \mathbb{Z}^+$ (i.e., both $a$ and $b$ are positive integers)

• The equation:

  $$a + b = \frac{a}{b} + \frac{b}{a}$$

Question:

Find the value of:

$$a^2 + b^2 = \ ?$$

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Solution:
Simplify the expression to get:
ab(a+b) = a^2 + b^2
b.a^2 + a.b^2 = a^2 + b^2
=> a^2(b-1) + b^2(a-1) = 0
For LHS to be zero, only possibility is when a = b = 1
So answer = 2.

Q:

Let $\alpha, \beta$ be the roots of the equation:

$$x^2 + 5\sqrt{2}x + 10 = 0$$

with $\alpha > \beta$

Define a sequence:

$$P_n = \alpha^n - \beta^n,\quad n \in \mathbb{Z}^+$$

Evaluate the expression:

$$\frac{P_{17} \cdot P_{20} + 5\sqrt{2} \cdot P_{17} \cdot P_{19}}{P_{18} \cdot P_{19} + 5\sqrt{2} \cdot P_{18}^2}$$

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Solution:

Given the quadratic equation:

$$\alpha^2 + 5\sqrt{2} \alpha + 10 = 0$$

Multiply both sides by $\alpha$:

$$\alpha(\alpha + 5\sqrt{2}) = -10$$

So:

$$\alpha + 5\sqrt{2} = \frac{-10}{\alpha}$$

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Using the same idea for $\beta$, since it’s also a root of the equation:

$$\beta + 5\sqrt{2} = \frac{-10}{\beta}$$

$$\frac{P_{17} \cdot (\alpha^{20} - \beta^{20} + 5\sqrt{2}(\alpha^{19} - \beta^{19}))}{P_{18} \cdot (\alpha^{19} - \beta^{19} + 5\sqrt{2}(\alpha^{18} - \beta^{18}))}$$

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Using:

$$P_n = \alpha^n - \beta^n$$

So the expression becomes:

$$\frac{P_{17} \left( \alpha^{19}(\alpha + 5\sqrt{2}) - \beta^{19}(\beta + 5\sqrt{2}) \right)}{P_{18} \left( \alpha^{18}(\alpha + 5\sqrt{2}) - \beta^{18}(\beta + 5\sqrt{2}) \right)}$$

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Substituting:

$$\alpha + 5\sqrt{2} = \frac{-10}{\alpha}, \quad \beta + 5\sqrt{2} = \frac{-10}{\beta}$$

So:

$$= \frac{P_{17} \left( \alpha^{19} \cdot \frac{-10}{\alpha} - \beta^{19} \cdot \frac{-10}{\beta} \right)}{P_{18} \left( \alpha^{18} \cdot \frac{-10}{\alpha} - \beta^{18} \cdot \frac{-10}{\beta} \right)}$$

Simplifies to:

$$= \frac{P_{17} \left( -10\alpha^{18} + 10\beta^{18} \right)}{P_{18} \left( -10\alpha^{17} + 10\beta^{17} \right)} = \frac{P_{17} \cdot (-10 P_{18})}{P_{18} \cdot (-10 P_{17})} = 1$$