Solve-the-equation-in-R-5-x-1-1-x-x-2-2-x-1-4x-2-5x-5-

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

\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

Very nice solution \boldsymbol S er thank you

Commented by mathdanisur last updated on 30/Sep/21

But dear S er, we solve in R

But dear \boldsymbol S er, we solve in R

Commented by MJS_new last updated on 30/Sep/21

yes . as I said, no solution in R

yes . as I said, no solution in R

Commented by mathdanisur last updated on 30/Sep/21

THANKYOU DEAR SER

THANKYOU DEAR SER