Given-x-2-12-x-5-then-x-2-x-

Question Number 115544 by bemath last updated on 26/Sep/20

G i v e n x^{2} + 12 \sqrt{x} = 5

t h e n x + 2 \sqrt{x} ?

Commented by MJS_new last updated on 26/Sep/20

x^{2} + 12 \sqrt{x} = 5

has got 1 real and 2 complex solutions :

x = 3 - 2 \sqrt{2} \lor x = - 3 \pm 4 i

\Rightarrow x + 2 \sqrt{x} = 1 \lor x + 2 \sqrt{x} = - 1 \pm 8 i

Answered by bobhans last updated on 26/Sep/20

l e t \sqrt{x} = w; w ⩾ 0 \Rightarrow w^{4} + 12 w - 5 = 0

f a c t o r i n g

(w^{2} + 2 w - 1) (w^{2} - 2 w + 5) = 0

f o r w^{2} - 2 w + 5 > 0, \forall w \in R

f o r w^{2} + 2 w - 1 = 0

w^{2} + 2 w = 1 \Rightarrow {(\sqrt{x})}^{2} + 2 \sqrt{x} = 1

\Rightarrow x + 2 \sqrt{x} = 1

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