solve-for-x-a-R-x-2-ax-a-2-x-2-ax-a-2-1-

Question Number 119997 by behi83417@gmail.com last updated on 28/Oct/20

solve for x, a \in R .

\sqrt{x^{2} + a x + a^{2}} + \sqrt{x^{2} - a x + a^{2}} = 1

Answered by Lordose last updated on 28/Oct/20

x^{2} + ax + a^{2} = 1 - 2 \sqrt{x^{2} - ax + a^{2}} + x^{2} - {ax}^{2} + a^{2}

2 \sqrt{x^{2} - ax + a^{2}} = 1 - x^{2} - ax - a^{2} + x^{2} - ax + a^{2}

4 (x^{2} - ax + a^{2}) = {(1 - 2 ax)}^{2}

{4 x}^{2} - 4 ax + {4 a}^{2} = 1 - 4 ax + {4 a}^{2} x^{2}

(4 - {4 a}^{2}) x^{2} + {4 a}^{2} = 1

x^{2} = \frac{1 - {4 a}^{2}}{(4 - {4 a}^{2})}

x = \pm \frac{\sqrt{1 - {4 a}^{2}}}{2 \sqrt{1 - a^{2}}}

Commented by behi83417@gmail.com last updated on 28/Oct/20

thank you very much sir .

Answered by TANMAY PANACEA last updated on 28/Oct/20

x^{2} + a x + a^{2} = 1 + x^{2} - a x + a^{2} - 2 \sqrt{x^{2} - a x + a^{2}}

{(2 a x - 1)}^{2} = 4 (x^{2} - a x + a^{2})

4 a^{2} x^{2} - 4 a x + 1 - 4 x^{2} + 4 a x - 4 a^{2} = 0

4 x^{2} (a^{2} - 1) = 4 a^{2} - 1

x = \pm \sqrt{\frac{4 a^{2} - 1}{4 (a^{2} - 1)}}

Commented by behi83417@gmail.com last updated on 28/Oct/20

thanks in advance sir!

Commented by TANMAY PANACEA last updated on 28/Oct/20

m o s t w e l c o m e s i r

Answered by behi83417@gmail.com last updated on 28/Oct/20

x^{2} + ax + a^{2} = u^{2}, x^{2} - ax + a^{2} = v^{2}

\Rightarrow {\begin{cases} u + v = 1 \\ u^{2} - v^{2} = 2 ax \end{cases} \Rightarrow u - v = 2 ax

\Rightarrow {(u + v)}^{2} - 4 uv = {(2 ax)}^{2}

\Rightarrow uv = \frac{1 - {4 a}^{2} x^{2}}{4}, u + v = 1

\Rightarrow z^{2} - z + \frac{1 - {4 a}^{2} x^{2}}{4} = 0

\Rightarrow z = \frac{1 \pm \sqrt{1 - (1 - {4 a}^{2} x^{2})}}{2} = \frac{1 \pm 2 ax}{2}

\Rightarrow \sqrt{x^{2} + ax + a^{2}} = \frac{1 \pm 2 ax}{2} \Rightarrow

4 (x^{2} + ax + a^{2}) = 1 \pm 4 ax + {4 a}^{2} x^{2}

\Rightarrow {\begin{cases} 4 (1 - a^{2}) x^{2} = 1 - {4 a}^{2} \Rightarrow x = \pm \frac{1}{2} . \sqrt{\frac{1 - {4 a}^{2}}{1 - a^{2}}} \\ {4 x}^{2} + 4 ax + {4 a}^{2} = 1 - 4 ax + {4 a}^{2} x^{2} \Rightarrow \end{cases}

\Rightarrow 4 (1 - a^{2}) x^{2} + 8 ax + {4 a}^{2} - 1 = 0 \Rightarrow

\Rightarrow x = \frac{- 4 a \pm \sqrt{{16 a}^{2} - 4 (1 - a^{2}) ({4 a}^{2} - 1)}}{4 (1 - a^{2})} =

= \frac{- 2 a \pm \sqrt{{4 a}^{2} - ({4 a}^{2} - 1 - {4 a}^{4} + a^{2})}}{2 (1 - a^{2})} =

= \frac{2 a \pm \sqrt{{4 a}^{4} - a^{2} + 1}}{2 (a^{2} - 1)} .

Answered by MJS_new last updated on 29/Oct/20

for a = 0 we get x = \pm \frac{1}{2}

for a = \pm \frac{1}{2} we get x = 0

for a < - \frac{1}{2} \lor a > \frac{1}{2} we get no solution for x \in C

\Rightarrow - \frac{1}{2} ⩽ a ⩽ \frac{1}{2} \land - \frac{1}{2} ⩽ x ⩽ \frac{1}{2}

\Rightarrow let a = \frac{1}{2} \sin α \land α \in R

\Rightarrow x = \pm \frac{\cos α}{\sqrt{3 + \cos^{2} α}}

parametric solution is {\begin{cases} a = \frac{1}{2} \sin α \\ x = \pm \frac{\cos α}{\sqrt{3 + \cos^{2} α}} \end{cases} \land α \in R

or with t = \tan \frac{α}{2} {\begin{cases} a = \frac{t}{t^{2} + 1} \\ x = \pm \frac{t^{2} - 1}{2 \sqrt{t^{4} + t^{2} + 1}} \end{cases} \land t \in R

or solution is x = \pm \frac{\sqrt{1 - 4 a^{2}}}{2 \sqrt{1 - a^{2}}} \land - \frac{1}{2} ⩽ a ⩽ \frac{1}{2}

Commented by behi83417@gmail.com last updated on 29/Oct/20

thank you very much dear proph : MJS

please let me know what you think

about my answer .