Class X Session 2023-24 Question 27

MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041)

Question 27

\begin{aligned} i f, α β a r e Z e r o e s o f q u a d r a t i c p o l y n o m i a l 5 x^{2} + 5 x + 10 f i n d t h e v a l u e o f (1) α^{2} + β^{2}, (2) α^{-} 1 + β^{-} 1 \end{aligned}

Explanation:

$\begin{aligned} s i n c e α a n d β a r e t h e r o o t o f 5 x^{2} + 5 x + 10 a s w e k n o w t h a t i f q u a d r a t i c e q u a t i o n; \\ a x^{2} + b x + c = 0 t h e n S U M o f z e r o e s α + β = - \frac{b}{a}; a n d p r o d u c t o f Z e r o e s = \frac{c}{a} \\ s i n c e α a n d β a r e t h e r o o t o f 5 x^{2} + 5 x + 10 s i m i l a r l y f r o m t h e g i v e n e q u a t i o n α + β = \\ \frac{- 5}{5} a n d α β = \frac{1}{5}, n o w l e t s f i n d α^{2} + β^{2} \\ \Rightarrow (α + β)^{2} = α^{2} + β^{2} + 2 α β \\ o r α^{2} + β^{2} = (α + β)^{2} - 2 α β \\ n o w p u t t h e v a l u e s \\ α + β a n d α β \\ w e g e t α^{2} + β^{2} = (- 1)^{2} - 2 \times \frac{1}{5} \\ \Rightarrow 1 - \frac{2}{5} \\ \Rightarrow \frac{5 - 2}{5} \\ \Rightarrow \frac{3}{5} \\ N o w L e t s f i n d α^{- 1} + β^{- 1} = \frac{1}{α} + \frac{1}{β} A f t e r s i m p l i f y w e g e t \\ \Rightarrow \frac{α + β}{α β} n o w p u t t h e; v a l u e s f r o m a b o v e a n d w e g e t \\ \Rightarrow \frac{- 1}{\frac{1}{5}} \\ \Rightarrow \frac{- 5}{1} o r - 5 \end{aligned}$

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Class X Session 2023-24 Question 27

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