Find the real solution(s) of $\sqrt{3 - x} - \sqrt{x + 1} > \frac{1}{3}$ $sqrt(3-x)-sqrt(x+1) > 1/3$ ?

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Find the real solution(s) of $\sqrt{3 - x} - \sqrt{x + 1} > \frac{1}{3}$ $sqrt(3-x)-sqrt(x+1) > 1/3$ ?

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Answer 1 · 2016-09-04T17:17:39

$x \in [- 1, 1 - \frac{\sqrt{71}}{18})$ $x in [-1, 1 - sqrt(71)/18)$

Explanation:

First find values of $x$ $x$ where:

$\sqrt{3 - x} - \sqrt{x + 1} = \frac{1}{3}$ $sqrt(3-x)-sqrt(x+1) = 1/3$

Squaring both sides (which may introduce spurious solutions), we get:

$(3 - x) - 2 \sqrt{(3 - x) (x + 1)} + (x + 1) = \frac{1}{9}$ $(3-x)-2sqrt((3-x)(x+1))+(x+1) = 1/9$

which simplifies to:

$4 - 2 \sqrt{3 + 2 x - x^{2}} = \frac{1}{9}$ $4-2 sqrt(3+2x-x^2) = 1/9$

Multiply both sides by $9$ $9$ to get:

$36 - 18 \sqrt{3 + 2 x - x^{2}} = 1$ $36-18 sqrt(3+2x-x^2) = 1$

Hence:

$35 = 18 \sqrt{3 + 2 x - x^{2}}$ $35 = 18 sqrt(3+2x-x^2)$

Square both sides to get:

$1225 = 324 (3 + 2 x - x^{2}) = 972 + 648 x - 324 x^{2}$ $1225 = 324(3+2x-x^2) = 972+648x-324x^2$

Rearrange to get:

$324 x^{2} - 648 x + 253 = 0$ $324x^2-648x+253 = 0$

Divide through by $324$ $324$ to get:

$0 = x^{2} - 2 x + \frac{253}{324} = {(x - 1)}^{2} - \frac{71}{324}$ $0 = x^2-2x+253/324 = (x-1)^2-71/324$

Hence $x = 1 \pm \frac{\sqrt{71}}{18}$ $x = 1+-sqrt(71)/18$

Note that if $x = 1 + \frac{\sqrt{71}}{18}$ $x = 1+sqrt(71)/18$ then $3 - x < 2$ $3-x < 2$ , $x + 1 > 2$ $x+1 > 2$ and:

$\sqrt{3 - x} - \sqrt{x + 1} < 0$ $sqrt(3-x)-sqrt(x+1) < 0$

If $x = 1 - \frac{\sqrt{71}}{18}$ $x = 1-sqrt(71)/18$ then:

$\sqrt{3 - x} - \sqrt{x + 1} = \sqrt{2 + \frac{\sqrt{71}}{18}} - \sqrt{2 - \frac{\sqrt{71}}{18}} > 0$ $sqrt(3-x)-sqrt(x+1) = sqrt(2+sqrt(71)/18)-sqrt(2-sqrt(71)/18) > 0$

and we find:

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = {⎛ ⎝ \sqrt{2 + \frac{\sqrt{71}}{18}} - \sqrt{2 - \frac{\sqrt{71}}{18}} ⎞ ⎠}^{2}$ $(sqrt(3-x)-sqrt(x+1))^2 = (sqrt(2+sqrt(71)/18)-sqrt(2-sqrt(71)/18))^2$

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = (2 + \frac{\sqrt{71}}{18}) - 2 \sqrt{(2 + \frac{\sqrt{71}}{18}) (2 - \frac{\sqrt{71}}{18})} + (2 - \frac{\sqrt{71}}{18})$ $color(white)((sqrt(3-x)-sqrt(x+1))^2) = (2+sqrt(71)/18)-2sqrt((2+sqrt(71)/18)(2-sqrt(71)/18))+(2-sqrt(71)/18)$

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = 4 - 2 \sqrt{4 - \frac{71}{324}}$ $color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2sqrt(4-71/324)$

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = 4 - 2 \sqrt{\frac{1225}{324}}$ $color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2sqrt(1225/324)$

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = 4 - 2 \cdot \frac{35}{18}$ $color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2*35/18$

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = \frac{36}{9} - \frac{35}{9}$ $color(white)((sqrt(3-x)-sqrt(x+1))^2) = 36/9-35/9$

${(\sqrt{3 - x} - \sqrt{x + 1})}^{2} = \frac{1}{9}$ $color(white)((sqrt(3-x)-sqrt(x+1))^2) = 1/9$

So this is a valid solution.

Now if $- 1 \leq x < 1 - \frac{\sqrt{71}}{18}$ $-1 <= x < 1-sqrt(71)/18$ then both square roots are defined and $\sqrt{3 - x} - \sqrt{x + 1} > \frac{1}{3}$ $sqrt(3-x)-sqrt(x+1) > 1/3$

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