z>0 z^2 + 12(√z) = 5 z + 2(√z) = ?


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Question Number 149354 by mathdanisur last updated on 04/Aug/21

$z > 0$ $z^{2} + 12 \sqrt{z} = 5$ $z + 2 \sqrt{z} = ?$
Commented byRasheed.Sindhi last updated on 05/Aug/21

$⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣$ $GOOD S a y i n g :$ $♮ If you know nothing you must believe$ $everything . ε$ $⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢$
Commented byliberty last updated on 05/Aug/21
$(√z) = u⇒u^4 −12u−5=0 (u^2 +bu+5)(u^2 +au−1)=0 u^4 +au^3 −u^2 +bu^3 +abu^2 −bu+5u^2 +5au−5=0 u^4 +(a+b)u^3 +(ab+4)u^2 +(5a−b)u−5=0 a=−b ; 4=a^2 → { ((a=−2)),((b=2)) :} ⇔z^2 +12z−5=0 (z+2(√z)+5)(z−2(√z)−1)=0 { ((z+2(√z) =−5)),((z−2(√z) =1 )) :}$
$\sqrt{z} = u \Rightarrow u^{4} - 12 u - 5 = 0$ $(u^{2} + bu + 5) (u^{2} + au - 1) = 0$ $u^{4} + {au}^{3} - u^{2} + {bu}^{3} + {abu}^{2} - bu + {5 u}^{2} + 5 au - 5 = 0$ $u^{4} + (a + b) u^{3} + (ab + 4) u^{2} + (5 a - b) u - 5 = 0$ $a = - b; 4 = a^{2} \to {\begin{cases} a = - 2 \\ b = 2 \end{cases}$ $\Leftrightarrow z^{2} + 12 z - 5 = 0$ $(z + 2 \sqrt{z} + 5) (z - 2 \sqrt{z} - 1) = 0$ ${\begin{cases} z + 2 \sqrt{z} = - 5 \\ z - 2 \sqrt{z} = 1 \end{cases}$
Commented bymathdanisur last updated on 05/Aug/21

$T h a n k Y o u S e r$
Commented byMJS_new last updated on 05/Aug/21
$if you know nothing you must believe everything this is methodically wrong you show me how z+2(√z)=−5 can be solved z>0∧(√z)=u ⇒ u>0 u^4 +12u−5=0 (u^2 −au−b)(u^2 +au−c)=0 ⇒ { ((a^2 +b+c=0)),((a(c−b)−12=0)),((bc+5=0)) :} ⇔ a=2∧b=−5∧c=1 (u^2 −2u+5)(u^2 +2u−1)=0 ⇒ u=−1±(√2)∨u=1±2i u>0 ⇒ u=−1+(√2)=(√z) ⇒ z=3−2(√2) ⇒ z+2(√z)=1 if we drop z>0 z=u^2 z=3±2(√2)∨z=−3±4i but z=3+2(√2) doesn′t solve the given equation ⇒ z=3−2(√2)∧z+2(√z)=1 ∨ z=−3±4i∧z+2(√z)=−1±8i$
$if you know nothing you must believe$ $everything$ $this is methodically wrong$ $you show me how z + 2 \sqrt{z} = - 5 can be solved$ $z > 0 \land \sqrt{z} = u \Rightarrow u > 0$ $u^{4} + 12 u - 5 = 0$ $(u^{2} - a u - b) (u^{2} + a u - c) = 0$ $\Rightarrow$ ${\begin{cases} a^{2} + b + c = 0 \\ a (c - b) - 12 = 0 \\ b c + 5 = 0 \end{cases} \Leftrightarrow a = 2 \land b = - 5 \land c = 1$ $(u^{2} - 2 u + 5) (u^{2} + 2 u - 1) = 0$ $\Rightarrow$ $u = - 1 \pm \sqrt{2} \lor u = 1 \pm 2 i$ $u > 0 \Rightarrow u = - 1 + \sqrt{2} = \sqrt{z} \Rightarrow z = 3 - 2 \sqrt{2}$ $\Rightarrow$ $z + 2 \sqrt{z} = 1$ $if we drop z > 0$ $z = u^{2}$ $z = 3 \pm 2 \sqrt{2} \lor z = - 3 \pm 4 i$ $but z = 3 + 2 \sqrt{2} {doesn}^{'} t solve the given equation$ $\Rightarrow$ $z = 3 - 2 \sqrt{2} \land z + 2 \sqrt{z} = 1$ $\lor$ $z = - 3 \pm 4 i \land z + 2 \sqrt{z} = - 1 \pm 8 i$
Commented bymathdanisur last updated on 06/Aug/21

$Thank you Ser, cool$
	Answered by MJS_new last updated on 04/Aug/21

	$z = 3 - 2 \sqrt{2} \Rightarrow answer is 1$
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