If alpha and bita are the zeros of quadratic polynomial x^2 - 5 then form a quadratic polynomial whose zeros are 1+alpha and 1+bita

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Solution

Dear student,
$L e t p (x) = x^{2} - 5 \Rightarrow x^{2} - {(\sqrt{5})}^{2} = 0 \Rightarrow (x + \sqrt{5}) (x - \sqrt{5}) = 0 t h u s t h e r o o t s a r e - \sqrt{5}, \sqrt{5} \sin c e i t i s g i v e n α a n d β a r e t h e r o o t s o f p (x) s o t a k i n g α = - \sqrt{5} a n d β = \sqrt{5} t h e n 1 + α = 1 - \sqrt{5} a n d 1 + β = 1 + \sqrt{5} G e n e r a l f o r m o f q u a d r a t i c e q u a t i o n c a n b e g i v e n b y x^{2} - (s u m o f z e r o e s) x + (p r o d u c t o f z e r o e s) = 0 \Rightarrow x^{2} - (1 - \sqrt{5} + 1 + \sqrt{5}) x + (1 - \sqrt{5}) (1 + \sqrt{5}) = 0 \Rightarrow x^{2} - 2 x + (1 - 5) = 0 \Rightarrow x^{2} - 2 x - 4 = 0$
Regards

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