x,y,z are real numbers { ((x+y+z=4)),((x^2 +y^2 +z^2 =10)),((x^3 +y^3 +z^3 =22)) :} Find x^4 +y^4 +z^4


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Question Number 120952 by abony1303 last updated on 04/Nov/20
$x,y,z are real numbers { ((x+y+z=4)),((x^2 +y^2 +z^2 =10)),((x^3 +y^3 +z^3 =22)) :} Find x^4 +y^4 +z^4$
$x, y, z a r e r e a l n u m b e r s$ ${\begin{cases} x + y + z = 4 \\ x^{2} + y^{2} + z^{2} = 10 \\ x^{3} + y^{3} + z^{3} = 22 \end{cases} F i n d x^{4} + y^{4} + z^{4}$
	Answered by MJS_new last updated on 04/Nov/20
	$z=4−x−y let x=u−(√v)∧y=u+(√v) ⇒ { ((z=4−2u)),((2v+2(3u^2 −8u+3)=0)),((6uv−6(u^3 −8u^2 +16u−7)=0)) :} ⇒ −3u^2 −8u+3=((u^3 −8u^2 +16u−7)/u) ⇔ u^3 −4u^2 +((19)/4)u−(7/4)=0 ⇔ (u−1)(u−((3−(√2))/2))(u−((3+(√2))/2))=0 take u=1 [or any other of the solutions] ⇒ v=2 ⇒ x=1−(√2)∧y=1+(√2)∧z=2 ⇒ answer is 50$
	$z = 4 - x - y$ $let x = u - \sqrt{v} \land y = u + \sqrt{v}$ $\Rightarrow$ ${\begin{cases} z = 4 - 2 u \\ 2 v + 2 (3 u^{2} - 8 u + 3) = 0 \\ 6 u v - 6 (u^{3} - 8 u^{2} + 16 u - 7) = 0 \end{cases}$ $\Rightarrow - 3 u^{2} - 8 u + 3 = \frac{u^{3} - 8 u^{2} + 16 u - 7}{u}$ $\Leftrightarrow u^{3} - 4 u^{2} + \frac{19}{4} u - \frac{7}{4} = 0$ $\Leftrightarrow (u - 1) (u - \frac{3 - \sqrt{2}}{2}) (u - \frac{3 + \sqrt{2}}{2}) = 0$ $take u = 1 [or any other of the solutions]$ $\Rightarrow v = 2$ $\Rightarrow x = 1 - \sqrt{2} \land y = 1 + \sqrt{2} \land z = 2$ $\Rightarrow answer is 50$
	Answered by Jamshidbek2311 last updated on 04/Nov/20

	$x y + y z + x z = 3$ $x y z = - 2$ ${(x y)}^{2} + {(y z)}^{2} + {(x z)}^{2} = 25$ $x^{4} + y^{4} + z^{4} = 50$
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