x-3-2-x-4-x-5-6-x-2-2-x-

Question Number 149041 by mathdanisur last updated on 02/Aug/21

\sqrt{x - 3 + 2 \sqrt{x - 4}} - \sqrt{x + 5 - 6 \sqrt{x - 2}} = 2

\Rightarrow x = ?

Commented by bramlexs22 last updated on 02/Aug/21

Commented by MJS_new last updated on 02/Aug/21

your picture is not enough . do they intersect

at x \approx 4.5 or not ? also there might be an

intersection for x ⩾ 13 + 6 \sqrt{2}

Commented by bramlexs22 last updated on 02/Aug/21

the curve y = \sqrt{x - 3 + 2 \sqrt{x - 4}} - \sqrt{x + 5 - 6 \sqrt{x - 2}}

and the line y = 2 does not intersection

Answered by MJS_new last updated on 02/Aug/21

f_{1} (x) = \sqrt{x - 3 + 2 \sqrt{x - 4}}

defined for x ⩾ 4

strictly increasing for x ⩾ 4

f_{2} (x) = \sqrt{x + 5 - 6 \sqrt{x - 2}}

defined for 2 ⩽ x ⩽ 13 - 6 \sqrt{2} \lor x ⩾ 13 + 6 \sqrt{2}

strictly decreasing for 2 ⩽ x ⩽ 13 - 6 \sqrt{2}

strictly increasing for x ⩾ 13 + 6 \sqrt{2}

for 4 ⩽ x ⩽ 13 - 6 \sqrt{2} the maximum of

f_{1} (x) - f_{2} (x) = \sqrt{10 - 6 \sqrt{2} - 2 \sqrt{3} - 2 \sqrt{6}} \approx 1.72 < 2

\Rightarrow no solution

for x ⩾ 13 + 2 \sqrt{6} : 4 < f_{1} (x) - f_{2} (x) ⩽ \sqrt{10 + 6 \sqrt{2} + 2 \sqrt{3} + 2 \sqrt{6}} \approx 5.18

\Rightarrow no solution

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