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Question Number 144922 by mathdanisur last updated on 30/Jun/21

if z^2 - 16(√z) = 12 find z - 2(√z) = ?

ifz2−16z=12findz−2z=?

Answered by liberty last updated on 30/Jun/21

x^4 −16x−12=0 x^4 −16x−12=(x^2 +ax+b)(x^2 +cx+d) (x^2 +ax+b)(x^2 +cx+d)= x^4 +(a+c)x^3 +(b+d+ac)x^2 +(ad+bc)x+bd { ((bd=−12)),((a+c=0⇒c=−a)),((b+d+ac=0)),((ac+bd=−16)) :} ⇔ ac =−4⇒−a^2 =−4 ; a=±2 & c = ∓ 2 case(1) { ((a=2)),((c=−2)) :}⇒ b+d=4 ∧ bd =−12 ⇒b(4−b)=−12 ⇒b^2 −4b−12=0 ; (b−6)(b+2)=0 → { ((b=6⇒d=−2)),((b=−2⇒d=−6)) :} we get x^4 −16x−12=(x^2 +2x+6)(x^2 −2x−2)=0 we want compute z−2(√z) = x^2 −2x = 2 case(2) x^4 −16x−12=(x^2 +2x−2)(x^2 −2x−6) we want compute z−2(√z) = x^2 −2x = 6

x4−16x−12=0x4−16x−12=(x2+ax+b)(x2+cx+d)(x2+ax+b)(x2+cx+d)=x4+(a+c)x3+(b+d+ac)x2+(ad+bc)x+bd{bd=−12a+c=0⇒c=−ab+d+ac=0ac+bd=−16⇔ac=−4⇒−a2=−4;a=±2&c=∓2case(1){a=2c=−2⇒b+d=4∧bd=−12⇒b(4−b)=−12⇒b2−4b−12=0;(b−6)(b+2)=0→{b=6⇒d=−2b=−2⇒d=−6wegetx4−16x−12=(x2+2x+6)(x2−2x−2)=0wewantcomputez−2z=x2−2x=2case(2)x4−16x−12=(x2+2x−2)(x2−2x−6)wewantcomputez−2z=x2−2x=6

Commented by mathdanisur last updated on 30/Jun/21

thank you Sir

thankyouSir

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