Let alpha and beta be the roots of x^2+sqrt{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the equation $x^2+\sqrt{3}x-16=0$,since $\alpha$ and $\beta$ are roots,we have $\alpha^2+\sqrt{3}\alpha-16=0$ and $\beta^2+\sqrt{3}\beta-16=0$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) For the equation $x^2+\sqrt{3}x-16=0$,since $\alpha$ and $\beta$ are roots,we have $\alpha^2+\sqrt{3}\alpha-16=0$ and $\beta^2+\sqrt{3}\beta-16=0$. Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively and adding,we get $P_n+\sqrt{3}P_{n-1}-16P_{n-2}=0$. For $n=25$,$P_{25}+\sqrt{3}P_{24}=16P_{23}$. Thus,$\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$. For the equation $x^2+3x-1=0$,since $\gamma$ and $\delta$ are roots,we have $\gamma^2+3\gamma-1=0$ and $\delta^2+3\delta-1=0$,which implies $\gamma^2-1=-3\gamma$ and $\delta^2-1=-3\delta$. Then $Q_{25}-Q_{23} = (\gamma^{25}+\delta^{25})-(\gamma^{23}+\delta^{23}) = \gamma^{23}(\gamma^2-1)+\delta^{23}(\delta^2-1) = \gamma^{23}(-3\gamma)+\delta^{23}(-3\delta) = -3(\gamma^{24}+\delta^{24}) = -3Q_{24}$. Thus,$\frac{Q_{25}-Q_{23}}{Q_{24}} = -3$. Finally,$8-3=5$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

Let $\alpha$ and $\beta$ be the roots of $x^2+\sqrt{3}x-16=0$,and $\gamma$ and $\delta$ be the roots of $x^2+3x-1=0$. If $P_{n}=\alpha^{n}+\beta^{n}$ and $Q_{n}=\gamma^{n}+\delta^{n}$,then $\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to

Similar Questions

Let $\alpha, \beta$ with $\alpha > \beta$ be the roots of the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$. Let $P_n = \alpha^n - \beta^n$ for $n \in \mathbb{N}$. Then $(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$ is equal to:

If $1, 2, 3$ and $4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$,then $a+2b+c$ is equal to

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x-6=0$ and if $a=\alpha^2+\beta^2+\gamma^2$,$b=\alpha\beta+\beta\gamma+\gamma\alpha$ and $c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$,then the correct inequality among the following is

Let $\alpha$ and $\beta$ be two distinct roots of the equation $x^3 + 3x^2 - 1 = 0$. The equation which has $(\alpha \beta)$ as its root is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module