find-the-solution-6-x-gt-x-4-

Question Number 117641 by bemath last updated on 13/Oct/20

find the solution \sqrt{6 - x} > x - 4

Answered by Don08q last updated on 13/Oct/20

(6 - x) > {(x - 4)}^{2} o r (6 - x) < {(x - 4)}^{2}

w h e n 6 - x > x^{2} - 8 x + 16

x^{2} - 7 x + 10 < 0

(x - 5) (x - 2) < 0

2 < x < 5

w h e n 6 - x < x^{2} - 8 x + 16

x^{2} - 7 x + 10 > 0

(x - 5) (x - 2) > 0

x < 2 \cup x > 5

H e n c e, {2 < x < 5 o r x < 2 \cup x > 5}

Commented by bemath last updated on 13/Oct/20

your solution 2 < x < 5 ?

i think it wrong . let substitute

x = - 3 \Rightarrow \sqrt{6 - (- 3)} > - 3 - 4

\Rightarrow 3 > - 7 it true

Answered by floor(10²Eta[1]) last updated on 13/Oct/20

x < 5

Answered by 1549442205PVT last updated on 13/Oct/20

\sqrt{6 - x} > x - 4 (1)

We need the condition x ⩽ 6 for the

root is defined . Then

i) For x < 4 then (1) is true

ii) For 4 ⩽ x ⩽ 6 (2) then

(1) \Leftrightarrow (6 - x) > {(x - 4)}^{2} = x^{2} - 8 x + 16

\Leftrightarrow x^{2} - 7 x + 10 < 0 \Leftrightarrow (x - 2) (x - 5) < 0

\Leftrightarrow 2 < x < 5 (3)

Combining (2) (3) we get 4 ⩽ x < 5

. Combining i) and ii) we get solutions

of given inequality are x \in (- \infty, 5)