Q_1 . A quadratic equation x^2 −3x+4=0 has roots α and β .without solving a)write down the values of α^2 +β^2 b) find the quadratic equation with integral coefficients,whose roots are (1/α^2 ) and(1/β^2 ) Q_2 . the first term of a GP is 32 and the sum to infinity is 48. find the common ratio and the 8^(th) term of the progression


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Question Number 35363 by Rio Mike last updated on 18/May/18

$Q_{1} . A q u a d r a t i c e q u a t i o n x^{2} - 3 x + 4 = 0$ $h a s r o o t s α a n d β . w i t h o u t s o l v i n g$ $a) w r i t e d o w n t h e v a l u e s o f α^{2} + β^{2}$ $b) f i n d t h e q u a d r a t i c e q u a t i o n$ $w i t h i n t e g r a l c o e f f i c i e n t s, w h o s e$ $r o o t s a r e \frac{1}{α^{2}} a n d \frac{1}{β^{2}}$ $Q_{2} . t h e f i r s t t e r m o f a G P i s 32$ $a n d t h e s u m t o i n f i n i t y i s 48 .$ $f i n d t h e c o m m o n r a t i o a n d t h e 8^{t h}$ $t e r m o f t h e p r o g r e s s i o n$
	Answered by Rasheed.Sindhi last updated on 18/May/18

	$Q_{1}$ $α & β are the roots of x^{2} - 3 x + 4 = 0$ $α + β = \frac{- (- 3)}{1} = 3, α β = \frac{4}{1} = 4$ $(a) α^{2} + β^{2} = {(α + β)}^{2} - 2 α β$ $= {(3)}^{2} - 2 (4) = 9 - 8 = 1$ $(b) Thwe equation whose roots are \frac{1}{α^{2}} & \frac{1}{β^{2}}$ $Sum - of - roots = \frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{α^{2} + β^{2}}{{(α β)}^{2}}$ $= \frac{1}{{(4)}^{2}} = 1 / 16$ $Product - of - roots = \frac{1}{α^{2}} \times \frac{1}{β^{2}} = \frac{1}{{(α β)}^{2}}$ $= \frac{1}{{(4)}^{2}} = \frac{1}{16}$ $The new equation :$ $x^{2} - (sum) x + product = 0$ $x^{2} - (1 / 16) x + (1 / 16) = 0$ ${16 x}^{2} - x + 1 = 0$
	Commented by MJS last updated on 18/May/18

	${you}^{'} re typing faster than me; -)$
	Commented by Rasheed.Sindhi last updated on 18/May/18

	$Only for this time . Always you beat me!$
	Answered by Rasheed.Sindhi last updated on 18/May/18

	$Q_{2}$ $a = 32$ $S_{\infty} = \frac{a}{1 - r} = 48$ $\frac{32}{1 - r} = 48$ $32 = 48 - 48 r$ $r = \frac{48 - 32}{48} = \frac{1}{3}$ $t_{n} = {a r}^{n - 1}$ $t_{8} = 32 \times {(1 / 3)}^{8 - 1} = \frac{32}{3^{7}} = \frac{32}{2187}$
	Answered by MJS last updated on 18/May/18

	$(x - α) (x - β) = x^{2} - (α + β) x + α β$ $we know that α β = 4 and α + β = 3 \Rightarrow$ $α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = 3^{2} - 2 \times 4 = 1$ $(x - \frac{1}{α^{2}}) (x - \frac{1}{β^{2}}) = x^{2} - \frac{α^{2} + β^{2}}{α^{2} β^{2}} x + \frac{1}{α^{2} β^{2}} =$ $= x^{2} - \frac{1}{16} x + \frac{1}{16}$ $16 x^{2} - x + 1 = 0$
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