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Question Number 155345 by mathdanisur last updated on 29/Sep/21

$Solve the equation in R$ $\frac{5 \sqrt{x + 1}}{\sqrt{1 - x + x^{2}} + 2 \sqrt{x + 1}} = {4 x}^{2} - 5 x + 5$
Commented by MJS_new last updated on 29/Sep/21

$no real solution$ $maximum of lhs \approx 1.86$ $minimum of rhs = \frac{55}{16}$
	Answered by ArielVyny last updated on 29/Sep/21
	$((5(√(x+1)))/( (√(1−x+x^2 ))+2(√(x+1))))=4x^2 −5x+5 { ((x≥−1)),((1−x+x^2 #4x+4→x^2 −5x−3#0)) :} ((5(√(x+1))((√(1−x+x^2 ))−2(√(x+1))))/((1−x+x^2 )−(4(x+1))))=4x^2 −5x+5 ((5(√(x+1))((√(1−x+x^2 ))−2(√(x+1))))/(x^2 −5x−3))=4x^2 −5x+5 5(√(x+1))((√(1−x+x^2 )))−10x−10=4x^2 −5x+5 5(√(x+1))((√(1−x+x^2 )))=4x^2 +5x+5 25(x+1)(1−x+x^2 )=(16x^4 +20x^3 +20x^2 )+(20x^3 +25x^2 +25x)+(20x^2 +25x+25) 16x^4 +15x^3 +65x^2 +50x=0 x(16x^3 +15^2 +65x+50)=0 x=0 ou 16x^3 +15x^2 +65x+50=0$
	$\frac{5 \sqrt{x + 1}}{\sqrt{1 - x + x^{2}} + 2 \sqrt{x + 1}} = 4 x^{2} - 5 x + 5$ $You can't use 'macro parameter character #' in math mode$ $\frac{5 \sqrt{x + 1} (\sqrt{1 - x + x^{2}} - 2 \sqrt{x + 1})}{(1 - x + x^{2}) - (4 (x + 1))} = 4 x^{2} - 5 x + 5$ $\frac{5 \sqrt{x + 1} (\sqrt{1 - x + x^{2}} - 2 \sqrt{x + 1})}{x^{2} - 5 x - 3} = 4 x^{2} - 5 x + 5$ $5 \sqrt{x + 1} (\sqrt{1 - x + x^{2}}) - 10 x - 10 = 4 x^{2} - 5 x + 5$ $5 \sqrt{x + 1} (\sqrt{1 - x + x^{2}}) = 4 x^{2} + 5 x + 5$ $25 (x + 1) (1 - x + x^{2}) = (16 x^{4} + 20 x^{3} + 20 x^{2}) + (20 x^{3} + 25 x^{2} + 25 x) + (20 x^{2} + 25 x + 25)$ $16 x^{4} + 15 x^{3} + 65 x^{2} + 50 x = 0$ $x (16 x^{3} + 15^{2} + 65 x + 50) = 0$ $x = 0 o u 16 x^{3} + 15 x^{2} + 65 x + 50 = 0$
	Commented by MJS_new last updated on 29/Sep/21

	$x = 0 is wrong . test it!$
	Commented by ArielVyny last updated on 29/Sep/21

	$y e s i h a v e s a y^{″} {o r}^{″} t o r e s p e c t a b = 0 {\begin{cases} a = 0 \\ b = 0 \end{cases} o r$
	Answered by MJS_new last updated on 29/Sep/21

	$\frac{5 \sqrt{x + 1}}{\sqrt{x^{2} - x + 1} + 2 \sqrt{x + 1}} = 4 x^{2} - 5 x + 5$ $- (8 x^{2} - 10 x + 5) \sqrt{x + 1} = (4 x^{2} - 5 x + 5) \sqrt{x^{2} - x + 1}$ $squaring (might introduce false solutions)$ $and teansforming$ $x^{2} (x^{4} - \frac{15}{2} x^{3} + \frac{217}{16} x^{2} - \frac{175}{16} x + \frac{15}{4}) = 0$ $obviously x = 0 is false$ $x^{4} - \frac{15}{2} x^{3} + \frac{217}{16} x^{2} - \frac{175}{16} x + \frac{15}{4} = 0$ $(x^{2} - \frac{25}{4} x + 5) (x^{2} - \frac{5}{4} x + \frac{3}{4}) = 0$ $x = \frac{25}{8} \pm \frac{\sqrt{305}}{8} [both false]$ $x = \frac{5}{8} \pm \frac{\sqrt{23}}{8} i [both true]$
	Commented by mathdanisur last updated on 29/Sep/21

	$Very nice solution S er thank you$
	Commented by mathdanisur last updated on 30/Sep/21

	$But dear S er, we solve in R$
	Commented by MJS_new last updated on 30/Sep/21

	$yes . as I said, no solution in R$
	Commented by mathdanisur last updated on 30/Sep/21

	$THANKYOU DEAR SER$
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