Solve for real number(((x−1)^2 )/2)+(((y−2)^2 )/4)+(((z−3)^2 )/6)+3=∣x−1∣+∣y−2∣+∣z−3∣


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Question Number 139577 by bemath last updated on 29/Apr/21

$Solve for real number \frac{{(x - 1)}^{2}}{2} + \frac{{(y - 2)}^{2}}{4} + \frac{{(z - 3)}^{2}}{6} + 3 =∣ x - 1 ∣ + ∣ y - 2 ∣ + ∣ z - 3 ∣$
	Answered by mr W last updated on 29/Apr/21

	$a =∣ x - 1 ∣⩾ 0, b =∣ y - 2 ∣⩾ 0, c =∣ z - 3 ∣⩾ 0$ $\frac{a^{2}}{2} + \frac{b^{2}}{4} + \frac{c^{2}}{6} + 3 = a + b + c$ $\frac{1}{2} (a^{2} - 2 a + 1) + \frac{1}{4} (b^{2} - 4 b + 4) + \frac{1}{6} (c^{2} - 6 c + 9) + 3 - \frac{1}{2} - 1 - \frac{3}{2} = 0$ $\frac{1}{2} {(a - 1)}^{2} + \frac{1}{4} {(b - 2)}^{2} + \frac{1}{6} {(c - 3)}^{2} = 0$ $\Rightarrow a =∣ x - 1 ∣= 1 \Rightarrow x = 2$ $\Rightarrow b =∣ y - 2 ∣= 2 \Rightarrow y = 4$ $\Rightarrow c =∣ z - 3 ∣= 3 \Rightarrow z = 6$
	Commented by Rasheed.Sindhi last updated on 29/Apr/21

	$V N i c e a s u s u a l!$
	Commented by mr W last updated on 29/Apr/21

	$t h a n k s s i r!$
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