{ (((√(x+y))+(√(x−y))=a)),((x^2 +y^2 =b [a,b∈R])) :}


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 65054 by behi83417@gmail.com last updated on 24/Jul/19
${ (((√(x+y))+(√(x−y))=a)),((x^2 +y^2 =b [a,b∈R])) :}$
${\begin{cases} \sqrt{x + y} + \sqrt{x - y} = a \\ x^{2} + y^{2} = b [a, b \in R] \end{cases}$
Commented by behi83417@gmail.com last updated on 24/Jul/19

$thanks in advance proph . Abdo .$ $[u + v = a \Rightarrow u^{2} + v^{2} + 2 uv = a^{2}] please check .$
Commented by mathmax by abdo last updated on 24/Jul/19

$y e s s i r . . .$
	Answered by MJS last updated on 24/Jul/19

	$x + y ⩾ 0 \Rightarrow x ⩾ - y$ $x - y ⩾ 0 \Rightarrow x ⩾ y$ $\Rightarrow 0 ⩽∣ y ∣⩽ x$ $\Rightarrow a ⩾ 0; b ⩾ 0$ $\sqrt{x + y} + \sqrt{x - y} = a \Rightarrow x = \frac{y^{2}}{a^{2}} + \frac{a^{2}}{4}$ $\Rightarrow a \neq 0$ $x^{2} + y^{2} = b$ $y^{4} + \frac{3 a^{4}}{2} y^{2} + \frac{a^{4} (a^{4} - 16 b)}{16} = 0$ $y^{2} = - \frac{3 a^{4}}{4} \pm \frac{a^{2} \sqrt{2 a^{4} + 4 b}}{2}$ $y^{2} ⩾ 0 \Rightarrow y^{2} = - \frac{3 a^{4}}{4} + \frac{a^{2} \sqrt{2 a^{4} + 4 b}}{2}$ $y = \pm \frac{a}{2} \sqrt{- 3 a^{2} + 2 \sqrt{2 a^{4} + 4 b}}$ $\Rightarrow x = - \frac{a^{2}}{2} + \frac{1}{2} \sqrt{2 a^{4} + 4 b}$
	Commented by MJS last updated on 24/Jul/19

	$if we allow y \in C$ $\Rightarrow y = \pm \frac{a}{2} \sqrt{- 3 a^{2} + 2 \sqrt{2 a^{4} + 4 b}} \lor y = \pm i \frac{a}{2} \sqrt{3 a^{2} + 2 \sqrt{2 a^{4} + 4 b}}$ $\Rightarrow x = - \frac{a^{2}}{2} + \frac{1}{2} \sqrt{2 a^{4} + 4 b} \lor x = - \frac{a^{2}}{2} - \frac{1}{2} \sqrt{2 a^{4} + 4 b}$
	Commented by MJS last updated on 24/Jul/19

	${you}^{'} re welcome, and thank you$
	Commented by behi83417@gmail.com last updated on 24/Jul/19

	$thank you very much proph : MJS .$ $you have a great place in this forum and$ $also for me . god bless you sir .$
	Answered by mr W last updated on 24/Jul/19
	$x+y=p x−y=q ⇒(√p)+(√q)=a>0 (x+y)^2 +(x−y)^2 =2(x^2 +y^2 ) ⇒p^2 +q^2 =2b let P=(√p) ,Q=(√q) P+Q=a⇒Q=a−P P^4 +Q^4 =2b⇒P^4 +(a−P)^4 =2b ⇒2P^4 −4aP^3 +6a^2 P^2 −4a^3 P+a^4 −2b=0 let λ=(P/a) ⇒λ^4 −2λ^3 +3λ^2 −2λ+((1/2)−(b/a^4 ))=0 ⇒λ^4 −2λ^3 +3λ^2 −2λ+δ=0 ⇒λ_(1,2) =(1/2)[1±(√(4(√(1−δ))−3))] ⇒λ_(2,3) =(1/2)[1±i(√(4(√(1−δ))+3))] P_(1,2) =aλ=(a/2)[1±(√(4(√(1−δ))−3))] Q_(1,2) =a−(a/2)[1±(√(4(√(1−δ))−3))]=(a/2)[1∓(√(4(√(1−δ))−3))] P_(3,4) =aλ=(a/2)[1±i(√(4(√(1−δ))+3))] Q_(3,4) =a−(a/2)[1±i(√(4(√(1−δ))+3))]=(a/2)[1∓i(√(4(√(1−δ))+3))] x=((p+q)/2)=((P^2 +Q^2 )/2) y=((p−q)/2)=((P^2 −Q^2 )/2) ⇒x_(1,2) =(a^2 /8){[1±(√(4(√(1−δ))−3))]^2 +[1∓(√(4(√(1−δ))−3))]^2 } ⇒x_(1,2) =(1/2)a^2 (2(√(1−δ))−1) ⇒x_(1,2) =(1/2)[(√(2(a^4 +2b)))−a^2 ] ⇒x_(3,4) =(a^2 /8){[1±i(√(4(√(1−δ))+3))]^2 +[1∓i(√(4(√(1−δ))+3))]^2 } ⇒x_(3,4) =−(a^2 /2)(2(√(1−δ))+1) ⇒x_(3,4) =−(1/2)[(√(2(a^4 +2b)))+a^2 ] ⇒y_(1,2) =(a^2 /8){[1±(√(4(√(1−δ))−3))]^2 −[1∓(√(4(√(1−δ))−3))]^2 } ⇒y_(1,2) =±(a^2 /2)(√(4(√(1−δ))−3)) ⇒y_(1,2) =±(a/2)(√(2(√(2(a^4 +2b)))−3a^2 )) ⇒y_(3,4) =(a^2 /8){[1±i(√(4(√(1−δ))+3))]^2 −[1∓i(√(4(√(1−δ))+3))]^2 } ⇒y_(3,4) =±((a^2 i)/2)(√(4(√(1−δ))+3)) ⇒y_(3,4) =±((ai)/2)(√(2(√(2(a^4 +2b)))+3a^2 ))$
	$x + y = p$ $x - y = q$ $\Rightarrow \sqrt{p} + \sqrt{q} = a > 0$ ${(x + y)}^{2} + {(x - y)}^{2} = 2 (x^{2} + y^{2})$ $\Rightarrow p^{2} + q^{2} = 2 b$ $l e t P = \sqrt{p}, Q = \sqrt{q}$ $P + Q = a \Rightarrow Q = a - P$ $P^{4} + Q^{4} = 2 b \Rightarrow P^{4} + {(a - P)}^{4} = 2 b$ $\Rightarrow 2 P^{4} - 4 {a P}^{3} + 6 a^{2} P^{2} - 4 a^{3} P + a^{4} - 2 b = 0$ $l e t λ = \frac{P}{a}$ $\Rightarrow λ^{4} - 2 λ^{3} + 3 λ^{2} - 2 λ + (\frac{1}{2} - \frac{b}{a^{4}}) = 0$ $\Rightarrow λ^{4} - 2 λ^{3} + 3 λ^{2} - 2 λ + δ = 0$ $\Rightarrow λ_{1, 2} = \frac{1}{2} [1 \pm \sqrt{4 \sqrt{1 - δ} - 3}]$ $\Rightarrow λ_{2, 3} = \frac{1}{2} [1 \pm i \sqrt{4 \sqrt{1 - δ} + 3}]$ $P_{1, 2} = a λ = \frac{a}{2} [1 \pm \sqrt{4 \sqrt{1 - δ} - 3}]$ $Q_{1, 2} = a - \frac{a}{2} [1 \pm \sqrt{4 \sqrt{1 - δ} - 3}] = \frac{a}{2} [1 \mp \sqrt{4 \sqrt{1 - δ} - 3}]$ $P_{3, 4} = a λ = \frac{a}{2} [1 \pm i \sqrt{4 \sqrt{1 - δ} + 3}]$ $Q_{3, 4} = a - \frac{a}{2} [1 \pm i \sqrt{4 \sqrt{1 - δ} + 3}] = \frac{a}{2} [1 \mp i \sqrt{4 \sqrt{1 - δ} + 3}]$ $x = \frac{p + q}{2} = \frac{P^{2} + Q^{2}}{2}$ $y = \frac{p - q}{2} = \frac{P^{2} - Q^{2}}{2}$ $\Rightarrow x_{1, 2} = \frac{a^{2}}{8} {{[1 \pm \sqrt{4 \sqrt{1 - δ} - 3}]}^{2} + {[1 \mp \sqrt{4 \sqrt{1 - δ} - 3}]}^{2}}$ $\Rightarrow x_{1, 2} = \frac{1}{2} a^{2} (2 \sqrt{1 - δ} - 1)$ $\Rightarrow x_{1, 2} = \frac{1}{2} [\sqrt{2 (a^{4} + 2 b)} - a^{2}]$ $\Rightarrow x_{3, 4} = \frac{a^{2}}{8} {{[1 \pm i \sqrt{4 \sqrt{1 - δ} + 3}]}^{2} + {[1 \mp i \sqrt{4 \sqrt{1 - δ} + 3}]}^{2}}$ $\Rightarrow x_{3, 4} = - \frac{a^{2}}{2} (2 \sqrt{1 - δ} + 1)$ $\Rightarrow x_{3, 4} = - \frac{1}{2} [\sqrt{2 (a^{4} + 2 b)} + a^{2}]$ $\Rightarrow y_{1, 2} = \frac{a^{2}}{8} {{[1 \pm \sqrt{4 \sqrt{1 - δ} - 3}]}^{2} - {[1 \mp \sqrt{4 \sqrt{1 - δ} - 3}]}^{2}}$ $\Rightarrow y_{1, 2} = \pm \frac{a^{2}}{2} \sqrt{4 \sqrt{1 - δ} - 3}$ $\Rightarrow y_{1, 2} = \pm \frac{a}{2} \sqrt{2 \sqrt{2 (a^{4} + 2 b)} - 3 a^{2}}$ $\Rightarrow y_{3, 4} = \frac{a^{2}}{8} {{[1 \pm i \sqrt{4 \sqrt{1 - δ} + 3}]}^{2} - {[1 \mp i \sqrt{4 \sqrt{1 - δ} + 3}]}^{2}}$ $\Rightarrow y_{3, 4} = \pm \frac{a^{2} i}{2} \sqrt{4 \sqrt{1 - δ} + 3}$ $\Rightarrow y_{3, 4} = \pm \frac{a i}{2} \sqrt{2 \sqrt{2 (a^{4} + 2 b)} + 3 a^{2}}$
	Commented by behi83417@gmail.com last updated on 24/Jul/19

	$my master is coming!$ $thanks in advance dear master .$ $your answers {don}^{'} t match with sir {MJS}^{'} s .$ $where is the problem sir ?$
	Commented by mr W last updated on 25/Jul/19

	$t h e y m a t c h s i r . m y a n s w e r j u s t a l s o$ $i n c l u d e s t h e c o m p l e x r o o t s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum