Resoudre dans R 1) a+b+c=2 a^2 +b^2 +c^2 =6 (1/a)+(1/b)+(1/c)=(1/2) 2) x^2 +xy+y^2 =3 y^2 +yz+z^2 =7 z^2 +zx+x^2 =13


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Question Number 180542 by a.lgnaoui last updated on 13/Nov/22

$R e s o u d r e d a n s R$ $1)$ $a + b + c = 2$ $a^{2} + b^{2} + c^{2} = 6$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}$ $2)$ $x^{2} + x y + y^{2} = 3$ $y^{2} + y z + z^{2} = 7$ $z^{2} + z x + x^{2} = 13$
	Answered by mr W last updated on 13/Nov/22

	$1)$ $a + b + c = 2$ ${(a + b + c)}^{2} = 2^{2} = 4$ $a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 4$ $\Rightarrow a b + b c + c a = \frac{4 - 6}{2} = - 1$ $\frac{a b + b c + c a}{a b c} = \frac{1}{2}$ $\Rightarrow a b c = 2 (- 1) = - 2$ $a, b, c a r e r o o t s o f$ $x^{3} - 2 x^{2} - x + 2 = 0$ $(x - 2) (x - 1) (x + 1) = 0$ $\Rightarrow a, b, c = (- 1, 1, 2)$
	Commented by a.lgnaoui last updated on 13/Nov/22

	$t h a n k s$
	Answered by mr W last updated on 14/Nov/22

	$2)$ $a = \sqrt{7}, b = \sqrt{13}, c = \sqrt{3}$ $Δ = \frac{\sqrt{(\sqrt{13} + \sqrt{7} + \sqrt{3}) (- \sqrt{13} + \sqrt{7} + \sqrt{3}) (\sqrt{13} - \sqrt{7} + \sqrt{3}) (\sqrt{13} + \sqrt{7} - \sqrt{3})}}{4} = \frac{5 \sqrt{3}}{4}$ $x = \pm \frac{4 \sqrt{6} \times \frac{5 \sqrt{3}}{4} + 3 \sqrt{2} (- 7 + 13 + 3)}{6 \sqrt{13 + 7 + 3 + 4 \sqrt{3} \times \frac{5 \sqrt{3}}{4}}}$ $x = \pm \frac{15 \sqrt{2} + 3 \sqrt{2} (- 7 + 13 + 3)}{6 \sqrt{13 + 7 + 3 + 15}} = \pm \frac{7}{\sqrt{19}}$ $y = \pm \frac{15 \sqrt{2} + 3 \sqrt{2} (7 - 13 + 3)}{6 \sqrt{13 + 7 + 3 + 15}} = \pm \frac{1}{\sqrt{19}}$ $z = \pm \frac{15 \sqrt{2} + 3 \sqrt{2} (7 + 13 - 3)}{6 \sqrt{13 + 7 + 3 + 15}} = \pm \frac{11}{\sqrt{19}}$
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