how-to-solve-c-c-2-c-6-c-2-2-lt-a-lt-3-4-lt-b-lt-3-c-a-b-2-

Question Number 30148 by .none. last updated on 17/Feb/18

h o w t o s o l v e \sqrt{{(c - \sqrt{- c})}^{2}} + \sqrt{{(c + 6 - \sqrt{- c})}^{2}}

2 < a < 3, - 4 < b < - 3, c = \frac{a + b}{2}

Answered by mrW2 last updated on 17/Feb/18

- 2 < a + b < 0

- 1 < c = \frac{a + b}{2} < 0

l e t d = - c

0 < d < 1

\sqrt{{(c - \sqrt{- c})}^{2}} + \sqrt{{(c + 6 - \sqrt{- c})}^{2}}

= \sqrt{{(- d - \sqrt{d})}^{2}} + \sqrt{{(- d + 6 - \sqrt{d})}^{2}}

= \sqrt{{(d + \sqrt{d})}^{2}} + \sqrt{{(6 - d - \sqrt{d})}^{2}}

= d + \sqrt{d} + 6 - d - \sqrt{d}

= 6

Commented by MJS last updated on 17/Feb/18

you had been slightly faster than me; -)

Answered by MJS last updated on 17/Feb/18

\Rightarrow - 1 ≦ c < 0

\Rightarrow \sqrt{{(c - \sqrt{- c})}^{2}} =∣ c - \sqrt{- c} ∣= \sqrt{- c} - c

\sqrt{{(c + 6 - \sqrt{- c})}^{2}} =∣ c + 6 - \sqrt{- c} ∣= 6 - (\sqrt{- c} - c)

\Rightarrow the solution is 6

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