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Question Number 194172 by sonukgindia last updated on 29/Jun/23

Commented by Frix last updated on 29/Jun/23
${ ((x^2 +y^2 =5)),((x^4 +y^4 =55)) :} x=±(√(a−(√b)))∧y=±(√(a+(√b))) { ((2a=5)),((2a^2 +2b=55)) :} a=(5/2)∧b=((85)/4) x=±((√(−10+2(√(85))))/2)i∧y=±((√(10+2(√(85))))/2) x+y=±((√(10+2(√(85))))/2)±((√(−10+2(√(85))))/2)i [4 solutions]$
${\begin{cases} x^{2} + y^{2} = 5 \\ x^{4} + y^{4} = 55 \end{cases}$ $x = \pm \sqrt{a - \sqrt{b}} \land y = \pm \sqrt{a + \sqrt{b}}$ ${\begin{cases} 2 a = 5 \\ 2 a^{2} + 2 b = 55 \end{cases}$ $a = \frac{5}{2} \land b = \frac{85}{4}$ $x = \pm \frac{\sqrt{- 10 + 2 \sqrt{85}}}{2} i \land y = \pm \frac{\sqrt{10 + 2 \sqrt{85}}}{2}$ $x + y = \pm \frac{\sqrt{10 + 2 \sqrt{85}}}{2} \pm \frac{\sqrt{- 10 + 2 \sqrt{85}}}{2} i$ $[4 solutions]$
	Answered by cortano12 last updated on 29/Jun/23

	$Let x + y = m$ $x^{2} + y^{2} + 2 xy = m^{2}$ $2 xy = m^{2} - 5$ ${(x^{2} + y^{2})}^{2} \overset{}{} = 25$ $x^{4} + y^{4} + 2 {(xy)}^{2} = 25$ $55 + 2 {(\frac{m^{2} - 5}{2})}^{2} = 25$
	Answered by behi834171 last updated on 30/Jun/23
	$x^4 +y^4 =(x^2 +y^2 )^2 −2(xy)^2 ⇒55=25−2(xy)^2 ⇒(xy)^2 =−15 (x+y)^2 =x^2 +y^2 −2(xy)=5±2i(√(15)) ⇒x+y=±(√(5±2i(√(15)))) [5+2i(√(15))]^(1/2) =a+ib⇒−40+20i(√(15))=a^2 −b^2 +2iab ⇒ { ((a^2 −b^2 =−40⇒a^2 −(((10(√(15)))/a))^2 =−40⇒)),((2ab=20(√(15)))) :} ⇒a^4 +40a^2 −1500=0 ⇒a^2 =((−40±(√(1600+6000)))/2)=−20±10(√(19)) ⇒a^2 =−(20+43.6),−(20−43.6)=−63.6,23.6 ⇒ { ((a=4.86⇒b=7.96)),((a=7.96i⇒b=−4.86i)) :} ⇒x+y=4.86+7.96i$
	$x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 {(x y)}^{2} \Rightarrow 55 = 25 - 2 {(x y)}^{2}$ $\Rightarrow {(x y)}^{2} = - 15$ ${(x + y)}^{2} = x^{2} + y^{2} - 2 (x y) = 5 \pm 2 i \sqrt{15}$ $\Rightarrow x + y = \pm \sqrt{5 \pm 2 i \sqrt{15}}$ ${[5 + 2 i \sqrt{15}]}^{\frac{1}{2}} = a + i b \Rightarrow - 40 + 20 i \sqrt{15} = a^{2} - b^{2} + 2 i a b$ $\Rightarrow {\begin{cases} a^{2} - b^{2} = - 40 \Rightarrow a^{2} - {(\frac{10 \sqrt{15}}{a})}^{2} = - 40 \Rightarrow \\ 2 a b = 20 \sqrt{15} \end{cases}$ $\Rightarrow a^{4} + 40 a^{2} - 1500 = 0$ $\Rightarrow a^{2} = \frac{- 40 \pm \sqrt{1600 + 6000}}{2} = - 20 \pm 10 \sqrt{19}$ $\Rightarrow a^{2} = - (20 + 43.6), - (20 - 43.6) = - 63.6, 23.6$ $\Rightarrow {\begin{cases} a = 4.86 \Rightarrow b = 7.96 \\ a = 7.96 i \Rightarrow b = - 4.86 i \end{cases}$ $\Rightarrow x + y = 4.86 + 7.96 i$
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