Relation between roots and coefficients Questions in English

(A) Given that $\alpha$ and $\beta$ are the roots of $x^{2}-px+1=0$.
By the relation between roots and coefficients,$\alpha+\beta=p$ and $\alpha\beta=1$.
Also,$\gamma$ is a root of $x^{2}+px+1=0$,so $\gamma^{2}+p\gamma+1=0$,which implies $\gamma^{2}=-p\gamma-1$.
Now,consider the expression $(\alpha+\gamma)(\beta+\gamma) = \alpha\beta + \gamma(\alpha+\beta) + \gamma^{2}$.
Substituting the known values:
$(\alpha+\gamma)(\beta+\gamma) = 1 + \gamma(p) + (-p\gamma-1)$.
$= 1 + p\gamma - p\gamma - 1 = 0$.

(A) Given that $\alpha$ and $\beta$ are the roots of $x^{2}-x-1=0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^{2}-\alpha-1=0 \implies \alpha^{2}=\alpha+1$
$\beta^{2}-\beta-1=0 \implies \beta^{2}=\beta+1$
Multiplying by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively:
$\alpha^{n+1}=\alpha^{n}+\alpha^{n-1}$
$\beta^{n+1}=\beta^{n}+\beta^{n-1}$
Adding these two equations:
$\alpha^{n+1}+\beta^{n+1}=(\alpha^{n}+\beta^{n})+(\alpha^{n-1}+\beta^{n-1})$
By definition of $S_{n}$,this is:
$S_{n+1}=S_{n}+S_{n-1}$
Thus,option $A$ is correct.

(D) Given the quadratic equation $x^{2}+p x+q=0$,the sum of roots $\alpha+\beta = -p$ and the product of roots $\alpha \beta = q$.
First,calculate $\alpha^{3}+\beta^{3}$:
$\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta) = (-p)^{3}-3 q(-p) = -p^{3}+3 p q = 3 p q-p^{3}$.
Next,calculate $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}$:
$\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2} - \alpha^{2} \beta^{2} = [(\alpha+\beta)^{2}-2 \alpha \beta]^{2} - (\alpha \beta)^{2} = [(-p)^{2}-2 q]^{2} - q^{2} = (p^{2}-2 q)^{2} - q^{2} = p^{4}-4 p^{2} q+4 q^{2}-q^{2} = p^{4}-4 p^{2} q+3 q^{2} = p^{2}(p^{2}-3 q)-q(p^{2}-3 q) = (p^{2}-q)(p^{2}-3 q)$.
Thus,the values are $3 p q-p^{3}$ and $(p^{2}-q)(p^{2}-3 q)$.

(D) Let $p(x) = ax^2 + bx + c$.
Given,the constant term $c = 1$.
Therefore,$p(x) = ax^2 + bx + 1$.
By the Remainder Theorem,$p(1) = 2$ and $p(-1) = 4$.
Substituting $x=1$: $a(1)^2 + b(1) + 1 = 2 \implies a + b = 1$ (Equation $I$).
Substituting $x=-1$: $a(-1)^2 + b(-1) + 1 = 4 \implies a - b = 3$ (Equation $II$).
Adding Equation $I$ and Equation $II$: $(a+b) + (a-b) = 1 + 3 \implies 2a = 4 \implies a = 2$.
Substituting $a=2$ into Equation $I$: $2 + b = 1 \implies b = -1$.
Thus,the polynomial is $p(x) = 2x^2 - x + 1$.
The sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is given by $-\frac{b}{a}$.
Sum of the roots $= -\frac{-1}{2} = \frac{1}{2}$.

(B) Given equation is $x^{2}+a x+b=0, (b \neq 0)$.
Its roots are $\alpha$ and $\beta$. Then,sum of roots $\alpha+\beta = -a$ and product of roots $\alpha \beta = b$.
Let the new roots be $S = \alpha-\frac{1}{\beta}$ and $T = \beta-\frac{1}{\alpha}$.
Sum of new roots $= S+T = (\alpha+\beta) - (\frac{1}{\alpha} + \frac{1}{\beta}) = (\alpha+\beta) - \frac{\alpha+\beta}{\alpha \beta} = -a - \frac{-a}{b} = -a + \frac{a}{b} = \frac{a(1-b)}{b}$.
Product of new roots $= ST = (\alpha-\frac{1}{\beta})(\beta-\frac{1}{\alpha}) = \alpha \beta - 1 - 1 + \frac{1}{\alpha \beta} = b - 2 + \frac{1}{b} = \frac{b^{2}-2b+1}{b} = \frac{(b-1)^{2}}{b}$.
The required quadratic equation is $x^{2} - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^{2} - \frac{a(1-b)}{b}x + \frac{(b-1)^{2}}{b} = 0$.
Multiplying by $b$,we get $b x^{2} - a(1-b)x + (b-1)^{2} = 0$.
Since $-a(1-b) = a(b-1)$,the equation is $b x^{2} + a(b-1)x + (b-1)^{2} = 0$.

(C) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $ax^{2}+bx+c=0$.
Therefore,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Given the condition $3b^{2} = 16ac$.
Dividing both sides by $a^{2}$,we get $3\left(\frac{b}{a}\right)^{2} = 16\left(\frac{c}{a}\right)$.
Substituting the values of sum and product of roots,we get $3(-\alpha-\beta)^{2} = 16\alpha\beta$.
$3(\alpha^{2}+\beta^{2}+2\alpha\beta) = 16\alpha\beta$.
$3\alpha^{2}+3\beta^{2}+6\alpha\beta = 16\alpha\beta$.
$3\alpha^{2}-10\alpha\beta+3\beta^{2} = 0$.
Dividing by $\beta^{2}$ (assuming $\beta \neq 0$),we get $3\left(\frac{\alpha}{\beta}\right)^{2}-10\left(\frac{\alpha}{\beta}\right)+3 = 0$.
Let $x = \frac{\alpha}{\beta}$,then $3x^{2}-10x+3 = 0$.
$3x^{2}-9x-x+3 = 0 \Rightarrow 3x(x-3)-1(x-3) = 0$.
$(3x-1)(x-3) = 0$.
So,$x = 3$ or $x = \frac{1}{3}$.
This implies $\frac{\alpha}{\beta} = 3$ or $\frac{\alpha}{\beta} = \frac{1}{3}$.
Therefore,$\alpha = 3\beta$ or $\beta = 3\alpha$.

(A) Given the quadratic equation $x^{2}-bx+c=0$ with roots $\sin \alpha$ and $\cos \alpha$.
From the relation between roots and coefficients:
$\sin \alpha + \cos \alpha = b$ $(i)$
$\sin \alpha \cdot \cos \alpha = c$ (ii)
We know that $(\sin \alpha + \cos \alpha)^{2} = \sin^{2} \alpha + \cos^{2} \alpha + 2 \sin \alpha \cos \alpha$.
Substituting the values: $b^{2} = 1 + 2c$,which implies $c = \frac{b^{2}-1}{2}$.
Since $-1 \leq \sin \alpha \leq 1$ and $-1 \leq \cos \alpha \leq 1$,we have $-\sqrt{2} \leq \sin \alpha + \cos \alpha \leq \sqrt{2}$,so $-\sqrt{2} \leq b \leq \sqrt{2}$.
Also,$\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2c$.
Since $-1 \leq \sin 2\alpha \leq 1$,we have $-1 \leq 2c \leq 1$,which implies $c \leq \frac{1}{2}$.

(A) Given that $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are roots of $x^{2}+px+q=0$.
Sum of roots: $(\alpha+\sqrt{\beta}) + (\alpha-\sqrt{\beta}) = -p \Rightarrow 2\alpha = -p \Rightarrow \alpha = -\frac{p}{2}$.
Product of roots: $(\alpha+\sqrt{\beta})(\alpha-\sqrt{\beta}) = q \Rightarrow \alpha^{2}-\beta = q \Rightarrow \beta = \alpha^{2}-q = \frac{p^{2}}{4}-q$.
Thus,$p^{2}-4q = 4\beta$.
Substitute into the equation $(p^{2}-4q)(p^{2}x^{2}+4px)-16q=0$:
$4\beta(p^{2}x^{2}+4px) - 16q = 0$.
Since $p = -2\alpha$,$p^{2} = 4\alpha^{2}$ and $q = \alpha^{2}-\beta$:
$4\beta(4\alpha^{2}x^{2}-8\alpha x) - 16(\alpha^{2}-\beta) = 0$.
Divide by $4$:
$\beta(\alpha^{2}x^{2}-2\alpha x) - (\alpha^{2}-\beta) = 0$.
$\alpha^{2}\beta x^{2} - 2\alpha\beta x - \alpha^{2} + \beta = 0$.
Rearranging terms: $\alpha^{2}(\beta x^{2}-1) - \beta(2\alpha x - 1) = 0$ is not straightforward,so let's simplify $4\beta(p^{2}x^{2}+4px) = 16q$:
$4\beta(4\alpha^{2}x^{2}-8\alpha x) = 16(\alpha^{2}-\beta) \Rightarrow \beta(\alpha^{2}x^{2}-2\alpha x) = \alpha^{2}-\beta$.
$\alpha^{2}\beta x^{2} - 2\alpha\beta x + \beta = \alpha^{2} \Rightarrow \beta(\alpha x - 1)^{2} = \alpha^{2}$.
$(\alpha x - 1)^{2} = \frac{\alpha^{2}}{\beta} \Rightarrow \alpha x - 1 = \pm \frac{\alpha}{\sqrt{\beta}}$.
$\alpha x = 1 \pm \frac{\alpha}{\sqrt{\beta}} \Rightarrow x = \frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}$.

(B) Given that $\sin \theta$ and $\cos \theta$ are the roots of the quadratic equation $ax^2 - bx + c = 0$.
From the properties of roots,the sum of roots is $\sin \theta + \cos \theta = \frac{b}{a}$ and the product of roots is $\sin \theta \cdot \cos \theta = \frac{c}{a}$.
Squaring the sum of roots,we get $(\sin \theta + \cos \theta)^2 = (\frac{b}{a})^2$.
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{b^2}{a^2}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $1 + 2(\frac{c}{a}) = \frac{b^2}{a^2}$.
Multiplying by $a^2$,we get $a^2 + 2ac = b^2$.
Rearranging the terms,we obtain $a^2 - b^2 + 2ac = 0$.

(B) Let the roots of the equation $px^2+qx+r=0$ be $a\alpha$ and $b\alpha$.
From the relation between roots and coefficients:
Sum of roots: $a\alpha + b\alpha = (a+b)\alpha = -\frac{q}{p}$
Product of roots: $(a\alpha)(b\alpha) = ab\alpha^2 = \frac{r}{p}$
Now,consider the ratio:
$\frac{ab\alpha^2}{(a+b)^2\alpha^2} = \frac{ab}{(a+b)^2}$
Substituting the values:
$\frac{ab}{(a+b)^2} = \frac{r/p}{(-q/p)^2} = \frac{r}{p} \times \frac{p^2}{q^2} = \frac{pr}{q^2}$

(A) In a triangle $PQR$,we have $P + Q + R = \pi$. Since $\angle R = \pi / 2$,it follows that $P + Q = \pi / 2$,which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.
Given that $\tan(P/2)$ and $\tan(Q/2)$ are roots of $ax^2 + bx + c = 0$,by Vieta's formulas:
Sum of roots: $\tan(P/2) + \tan(Q/2) = -b/a$
Product of roots: $\tan(P/2) \tan(Q/2) = c/a$
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan(\frac{P}{2} + \frac{Q}{2}) = \tan(\frac{\pi}{4}) = 1$
$\frac{-b/a}{1 - c/a} = 1$
$\frac{-b}{a - c} = 1$
$-b = a - c$
$c = a + b$

(A) Given the quadratic equation $\lambda x^{2} - (\lambda + 3)x + 3 = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = \frac{\lambda + 3}{\lambda}$ and $\alpha \beta = \frac{3}{\lambda}$.
Given $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$,which implies $\frac{\beta - \alpha}{\alpha \beta} = \frac{1}{3}$.
Since $\alpha < \beta$,$\beta - \alpha > 0$. Thus,$\beta - \alpha = \frac{\alpha \beta}{3} = \frac{3/\lambda}{3} = \frac{1}{\lambda}$.
Using the identity $(\beta - \alpha)^{2} = (\alpha + \beta)^{2} - 4\alpha \beta$,we get:
$(\frac{1}{\lambda})^{2} = (\frac{\lambda + 3}{\lambda})^{2} - 4(\frac{3}{\lambda})$.
$\frac{1}{\lambda^{2}} = \frac{\lambda^{2} + 6\lambda + 9}{\lambda^{2}} - \frac{12}{\lambda}$.
Multiplying by $\lambda^{2}$ (where $\lambda \neq 0$):
$1 = \lambda^{2} + 6\lambda + 9 - 12\lambda$.
$\lambda^{2} - 6\lambda + 8 = 0$.
$(\lambda - 2)(\lambda - 4) = 0$.
So,$\lambda = 2$ or $\lambda = 4$.
The sum of all possible values of $\lambda$ is $2 + 4 = 6$.

(B) Given $\beta - \alpha = \sqrt{11}i$ and $\beta^2 - \alpha^2 = 3\sqrt{11}i$.
Since $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$,we have $3\sqrt{11}i = (\sqrt{11}i)(\beta + \alpha)$,which implies $\beta + \alpha = 3$.
We know that $\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha^2 + \alpha\beta)$.
From $(\beta + \alpha)^2 = 9$,we have $\beta^2 + \alpha^2 + 2\alpha\beta = 9$.
From $(\beta - \alpha)^2 = -11$,we have $\beta^2 + \alpha^2 - 2\alpha\beta = -11$.
Subtracting the two equations: $4\alpha\beta = 20$,so $\alpha\beta = 5$.
Then $\beta^2 + \alpha^2 = 9 - 2(5) = -1$.
Substituting these into the expression for $\beta^3 - \alpha^3$: $\beta^3 - \alpha^3 = (\sqrt{11}i)(-1 + 5) = 4\sqrt{11}i$.
Finally,$(\beta^3 - \alpha^3)^2 = (4\sqrt{11}i)^2 = 16 \times 11 \times (-1) = -176$.
Taking the absolute value or considering the magnitude as implied by the options,the result is $176$.