If 4−2(√5) and 4+2(√(5 )) are solutions of x^2 +(5a−b)x+(3b−a)=0 whete a and b are real numbers, determine the product of ab.


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Question Number 70874 by Mr. K last updated on 09/Oct/19

$I f 4 - 2 \sqrt{5} a n d 4 + 2 \sqrt{5} a r e s o l u t i o n s$ $o f x^{2} + (5 a - b) x + (3 b - a) = 0$ $w h e t e a a n d b a r e r e a l n u m b e r s,$ $d e t e r m i n e t h e p r o d u c t o f a b .$
	Answered by tw000001 last updated on 09/Oct/19
	$x=4±2(√5)→(x−4)^2 =20 →x^2 −8x−4=0 → { ((5a−b=−8)),((a−3b=4)) :} →(a,b)=(−2,−2) ∴ab=4$
	$x = 4 \pm 2 \sqrt{5} \to {(x - 4)}^{2} = 20$ $\to x^{2} - 8 x - 4 = 0$ $\to {\begin{cases} 5 a - b = - 8 \\ a - 3 b = 4 \end{cases}$ $\to (a, b) = (- 2, - 2)$ $∴ a b = 4$
	Answered by Rasheed.Sindhi last updated on 09/Oct/19

	$A n O t h e r W a y$ $S a y α, β a r e r o o t s .$ $α + β = 5 a - b = (4 - 2 \sqrt{5}) + (4 + 2 \sqrt{5}) = 8$ $α β = 3 b - a = (4 - 2 \sqrt{5}) \times (4 + 2 \sqrt{5})$ $= {(4)}^{2} - {(2 \sqrt{5})}^{2} = 16 - 20 = - 4$ $5 a - b = 8 \land 3 b - a = - 4$ $(a, b) = (- 2, - 2)$ $a b = (- 2) (- 2) = 4$
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