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Question-209633




Question Number 209633 by Abdullahrussell last updated on 17/Jul/24
Answered by som(math1967) last updated on 17/Jul/24
(x+1)(3x+2)(6x+5)^2 =6  ⇒(3x^2 +5x+2)(36x^2 +60x+25)=6  ⇒(a+2)(12a+25)=6  [let 3x^2 +5x=a]  ⇒12a^2 +49a+50−6=0  ⇒12a^2 +49a+44=0  ⇒12a^2 +16a+33a+44=0  ⇒4a(3a+4)+11(3a+4)=0  ⇒(3a+4)(4a+11)=0  (9x^2 +15x+4)(12x^2 +20x+11)=0  if (9x^2 +15x+4)=0   (3x)^2 +2.3x.(5/2)+((25)/4)=((25)/4)−4  ∴(3x+(5/2))^2 =(9/4)
(x+1)(3x+2)(6x+5)2=6(3x2+5x+2)(36x2+60x+25)=6(a+2)(12a+25)=6[let3x2+5x=a]12a2+49a+506=012a2+49a+44=012a2+16a+33a+44=04a(3a+4)+11(3a+4)=0(3a+4)(4a+11)=0(9x2+15x+4)(12x2+20x+11)=0if(9x2+15x+4)=0(3x)2+2.3x.52+254=2544(3x+52)2=94
Answered by MM42 last updated on 17/Jul/24
(1/6)[(6x+5)+1)(1/2)[(6x+5)−1)(6x+5)^2 =6  (u+1)(u−1)(u)^2 =72  (u^2 −1)u^2 −72=  u^4 −u^2 −72=0  (u^2 −9)(u^2 +8)=0  u=±3  &  u=±2(√2)i  (3x+(5/2))^2 =(1/4)(6x+5)^2 =(9/4)  ✓ &  −2 ✓
16[(6x+5)+1)12[(6x+5)1)(6x+5)2=6(u+1)(u1)(u)2=72(u21)u272=u4u272=0(u29)(u2+8)=0u=±3&u=±22i(3x+52)2=14(6x+5)2=94&2

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