solve for x,a∈R. (√(x^2 +ax+a^2 ))+(√(x^2 −ax+a^2 ))=1


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
	$x^2 +ax+a^2 =u^2 ,x^2 −ax+a^2 =v^2 ⇒ { ((u+v=1)),((u^2 −v^2 =2ax)) :}⇒u−v=2ax ⇒(u+v)^2 −4uv=(2ax)^2 ⇒uv=((1−4a^2 x^2 )/4),u+v=1 ⇒z^2 −z+((1−4a^2 x^2 )/4)=0 ⇒z=((1±(√(1−(1−4a^2 x^2 ))))/2)=((1±2ax)/2) ⇒(√(x^2 +ax+a^2 ))=((1±2ax)/2)⇒ 4(x^2 +ax+a^2 )=1±4ax+4a^2 x^2 ⇒ { ((4(1−a^2 )x^2 =1−4a^2 ⇒x=±(1/2).(√((1−4a^2 )/(1−a^2 ))))),((4x^2 +4ax+4a^2 =1−4ax+4a^2 x^2 ⇒)) :} ⇒4(1−a^2 )x^2 +8ax+4a^2 −1=0⇒ ⇒x=((−4a±(√(16a^2 −4(1−a^2 )(4a^2 −1))))/(4(1−a^2 )))= =((−2a±(√(4a^2 −(4a^2 −1−4a^4 +a^2 ))))/(2(1−a^2 )))= =((2a±(√(4a^4 −a^2 +1)))/(2(a^2 −1))) .■$
	$x^{2} + ax + a^{2} = u^{2}, x^{2} - ax + a^{2} = v^{2}$ $\Rightarrow {\begin{cases} u + v = 1 \\ u^{2} - v^{2} = 2 ax \end{cases} \Rightarrow u - v = 2 ax$ $\Rightarrow {(u + v)}^{2} - 4 uv = {(2 ax)}^{2}$ $\Rightarrow uv = \frac{1 - {4 a}^{2} x^{2}}{4}, u + v = 1$ $\Rightarrow z^{2} - z + \frac{1 - {4 a}^{2} x^{2}}{4} = 0$ $\Rightarrow z = \frac{1 \pm \sqrt{1 - (1 - {4 a}^{2} x^{2})}}{2} = \frac{1 \pm 2 ax}{2}$ $\Rightarrow \sqrt{x^{2} + ax + a^{2}} = \frac{1 \pm 2 ax}{2} \Rightarrow$ $4 (x^{2} + ax + a^{2}) = 1 \pm 4 ax + {4 a}^{2} x^{2}$ $\Rightarrow {\begin{cases} 4 (1 - a^{2}) x^{2} = 1 - {4 a}^{2} \Rightarrow x = \pm \frac{1}{2} . \sqrt{\frac{1 - {4 a}^{2}}{1 - a^{2}}} \\ {4 x}^{2} + 4 ax + {4 a}^{2} = 1 - 4 ax + {4 a}^{2} x^{2} \Rightarrow \end{cases}$ $\Rightarrow 4 (1 - a^{2}) x^{2} + 8 ax + {4 a}^{2} - 1 = 0 \Rightarrow$ $\Rightarrow x = \frac{- 4 a \pm \sqrt{{16 a}^{2} - 4 (1 - a^{2}) ({4 a}^{2} - 1)}}{4 (1 - a^{2})} =$ $= \frac{- 2 a \pm \sqrt{{4 a}^{2} - ({4 a}^{2} - 1 - {4 a}^{4} + a^{2})}}{2 (1 - a^{2})} =$ $= \frac{2 a \pm \sqrt{{4 a}^{4} - a^{2} + 1}}{2 (a^{2} - 1)} . ◼$
	Answered by MJS_new last updated on 29/Oct/20
	$for a=0 we get x=±(1/2) for a=±(1/2) we get x=0 for a<−(1/2)∨a>(1/2) we get no solution for x∈C ⇒ −(1/2)≤a≤(1/2)∧−(1/2)≤x≤(1/2) ⇒ let a=(1/2)sin α ∧α∈R ⇒ x=±((cos α)/( (√(3+cos^2 α)))) parametric solution is { ((a=(1/2)sin α)),((x=±((cos α)/( (√(3+cos^2 α)))))) :}∧α∈R or with t=tan (α/2) { ((a=(t/(t^2 +1)))),((x=±((t^2 −1)/(2(√(t^4 +t^2 +1)))))) :}∧t∈R or solution is x=±((√(1−4a^2 ))/(2(√(1−a^2 ))))∧−(1/2)≤a≤(1/2)$
	$for a = 0 we get x = \pm \frac{1}{2}$ $for a = \pm \frac{1}{2} we get x = 0$ $for a < - \frac{1}{2} \lor a > \frac{1}{2} we get no solution for x \in C$ $\Rightarrow - \frac{1}{2} ⩽ a ⩽ \frac{1}{2} \land - \frac{1}{2} ⩽ x ⩽ \frac{1}{2}$ $\Rightarrow let a = \frac{1}{2} \sin α \land α \in R$ $\Rightarrow x = \pm \frac{\cos α}{\sqrt{3 + \cos^{2} α}}$ $parametric solution is {\begin{cases} a = \frac{1}{2} \sin α \\ x = \pm \frac{\cos α}{\sqrt{3 + \cos^{2} α}} \end{cases} \land α \in R$ $or with t = \tan \frac{α}{2} {\begin{cases} a = \frac{t}{t^{2} + 1} \\ x = \pm \frac{t^{2} - 1}{2 \sqrt{t^{4} + t^{2} + 1}} \end{cases} \land t \in R$ $or solution is x = \pm \frac{\sqrt{1 - 4 a^{2}}}{2 \sqrt{1 - a^{2}}} \land - \frac{1}{2} ⩽ a ⩽ \frac{1}{2}$
	Commented by behi83417@gmail.com last updated on 29/Oct/20

	$thank you very much dear proph : MJS$ $please let me know what you think$ $about my answer .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum