Let α, β(α>β) be roots of quadratic equation x ^{2}- x -4=0 . If P

English

Gujarati

Hindi

4-2.Quadratic Equations and Inequations

hard

Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x ^{2}- x -4=0$. If $P _{ a }=\alpha^{ n }-\beta^{ n }, n \in N$, then $\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^{2}+ P _{14} P _{15}}{ P _{13} P _{14}}$ is equal to$......$

$15$

$14$

$13$

$16$

(JEE MAIN-2022)

Solution

$Pn =\alpha^{ n }-\beta^{ n } \quad x ^{2}- x -4=0$

$\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^{2}+ P _{14} P _{15}}{ P _{13} P _{14}}$

As $P _{ n }- P _{ n -1}=\left(\alpha^{ a }-\beta^{ n }\right)-\left(\alpha^{ n -1}-\beta^{ n -1}\right)$

$=\alpha^{ n -2}\left(\alpha^{2}-\alpha\right)-\beta^{ n -2}\left(\beta^{2}-\beta\right)$

$=4\left(\alpha^{ n -2}-\beta^{ n -2}\right)$

$P _{ a }- P _{ n -1}=4 P _{ n -2}$

Hence Expression $(1)$

$\frac{P_{16}\left(P_{15}-P_{14}\right)-P_{15}\left(P_{15}-P_{14}\right)}{P_{13} P_{14}}$

$=\frac{\left( P _{15}- P _{14}\right)\left( P _{16}- P _{15}\right)}{ P _{13} P _{14}}=\frac{\left(4 P _{13}\right)\left(4 P _{14}\right)}{ P _{13} P _{14}}=16$

Standard 11

Mathematics

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

normal

The roots of the equation ${x^4} – 2{x^3} + x = 380$ are

normal

Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^2-x-1=0$, where $\alpha \neq \beta$. For $n=0,1,2, \ldots$, let $a_n=$ $p \alpha^n+q \beta^n$.

$FACT$ : If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$.

($1$) $a_{12}=$

$[A]$ $a_{11}-a_{10}$ $[B]$ $a_{11}+a_{10}$ $[C]$ $2 a_{11}+a_{10}$ $[D]$ $a_{11}+2 a_{10}$

($2$) If $a_4=28$, then $p+2 q=$

$[A] 21$ $[B] 14$ $[C] 7$ $[D] 12$

answer the quetion ($1$) and ($2$)

medium

(IIT-2017)

The sum, of the squares of all the roots of the equation $x^2+|2 x-3|-4=0$, is :

hard

(JEE MAIN-2025)

Number of natural solutions of the equation $xyz = 2^5 \times 3^2 \times 5^2$ is equal to

normal

Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x ^{2}- x -4=0$. If $P _{ a }=\alpha^{ n }-\beta^{ n }, n \in N$, then $\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^{2}+ P _{14} P _{15}}{ P _{13} P _{14}}$ is equal to$......$

Solution

Similar Questions

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

The roots of the equation ${x^4} – 2{x^3} + x = 380$ are

The sum, of the squares of all the roots of the equation $x^2+|2 x-3|-4=0$, is :

Number of natural solutions of the equation $xyz = 2^5 \times 3^2 \times 5^2$ is equal to

Start a Free Trial Now