The quadratic polynomial p(x) has roots 1 and | vedclass

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(A) Let $p(x) = a(x-1)(x-\alpha) = a(x^2 - (1+\alpha)x + \alpha)$ and $q(x) = b(x-1)(x-\beta) = b(x^2 - (1+\beta)x + \beta)$.Given $r(x) = p(x) + q(x)

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$0$

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(A) Let $p(x) = a(x-1)(x-\alpha) = a(x^2 - (1+\alpha)x + \alpha)$ and $q(x) = b(x-1)(x-\beta) = b(x^2 - (1+\beta)x + \beta)$.Given $r(x) = p(x) + q(x) = (a+b)x^2 - (a(1+\alpha) + b(1+\beta))x + (a\alpha + b\beta)$.Since $\alpha$ and $\beta$ are roots of $r(x)$, the sum of roots $\alpha + \beta = \frac{a(1+\alpha) + b(1+\beta)}{a+b}$.$(a+b)(\alpha+\beta) = a + a\alpha + b + b\beta \implies a\alpha + b\beta = a + b$.Thus, $r(x) = (a+b)x^2 - (a+b+a\alpha+b\beta)x + (a+b) = (a+b)(x^2 - (1 + \frac{a\alpha+b\beta}{a+b})x + 1) = (a+b)(x^2 - 2x + 1) = (a+b)(x-1)^2$.For $p(x)$ and $q(x)$ to be quadratic, we assume $a, b 
eq 0$. As $x 	o \infty$, $\sqrt{p(x)} \approx \sqrt{a}x$ and $\sqrt{q(x)} \approx \sqrt{b}x$.For the limit to exist, we must have $a=b$. Let $a=b=k$. Then $p(x) = k(x-1)(x-\alpha)$ and $q(x) = k(x-1)(x-\beta)$.From $a\alpha + b\beta = a+b$, we get $k(\alpha+\beta) = 2k \implies \alpha+\beta = 2$.$\sqrt{p(x)} - \sqrt{q(x)} = \sqrt{k(x-1)} (\sqrt{x-\alpha} - \sqrt{x-\beta}) = \sqrt{k(x-1)} \frac{(x-\alpha) - (x-\beta)}{\sqrt{x-\alpha} + \sqrt{x-\beta}} = \sqrt{k(x-1)} \frac{\beta-\alpha}{\sqrt{x-\alpha} + \sqrt{x-\beta}}$.As $x 	o \infty$, this behaves as $\sqrt{k} \cdot x \cdot \frac{\beta-\alpha}{2\sqrt{x}} = \frac{\sqrt{k}(\beta-\alpha)}{2} \sqrt{x}$, which only converges if $\beta-\alpha = 0$, i.e., $\alpha=\beta=1$. However, if $\alpha=\beta=1$, $p(x)=q(x)$, so the limit is $0$.

The quadratic polynomial $p(x)$ has roots $1$ and $\alpha$, while quadratic polynomial $q(x)$ has roots $1$ and $\beta$. Let $\alpha$ and $\beta$ be the roots of $r(x) = p(x) + q(x)$. Then $\lim_{x \to \infty} [\sqrt{p(x)} - \sqrt{q(x)}] = $

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