Class X Session 2023-24 Question 27

 MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041)

Question 27

\begin{flalign} & if\;,\;\alpha\; \beta \; are \; Zeroes \; of \; quadratic\;polynomial\; 5x^2 + 5x + 10 \;find \; the\; value\; of\; (1) \;\alpha^2\;+ \beta^2 \; ,\;(2)\;\alpha^-{1} + \beta^-{1} &\ \end{flalign}

Explanation:  

\begin{flalign} & since\; \alpha\; and \; \beta\; are \;the \;root\; of\; 5x^2 + 5x + 10 as \;we \;know\; that\; if \; quadratic\; equation;&\\ & ax^2 + bx +c =0 \;then \; SUM \;of \; zeroes\; \alpha + \beta\ = -\frac{b}{a}; and \;product\; of \;Zeroes\; =\; \frac{c}{a}&\\ & since\; \alpha\; and \; \beta\; are \;the \;root\; of\;5x^2 + 5x + 10 \;similarly \;\ from\; the \;given\; equation\; \alpha + \beta = & \\ & \frac{-5}{5} and \; \alpha \beta \;= \frac{1}{5} , \; now \; lets \;find\; \alpha^2 + \beta^2 &\\ &\Rightarrow (\alpha + \beta)^2 = \alpha^2+\beta^2 + 2\alpha\beta &\\ & or\; \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta &\\ & now\; put\; the\; values\; &\\ & \alpha +\beta \; and \; \alpha\beta &\\ & we \; get\; \alpha^2 + \beta^2 = (-1)^2 - 2\times\frac{1}{5} &\\ & \Rightarrow 1 - \frac{2}{5} &\\ & \Rightarrow \frac{5-2}{5} &\\ & \Rightarrow \frac{3}{5} &\\ & Now \;Lets \;find\; \alpha^{-1} +\beta^{-1} = \frac{1}{\alpha}+\frac{1}{\beta} After\; simplify \; we\; get\ &\\ & \Rightarrow \frac{\alpha+\beta}{\alpha\beta} \; now \; put \; the ; values\; from \; above \; and\; we \; get\; &\\ & \Rightarrow \frac{-1}{\frac{1}{5}} &\\ & \Rightarrow \frac{-5}{1} \; or\; -5 \end{flalign}