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-

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 \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