Recent content by emrys

My bad. it's 5x2

we will do it again from beginning we have (x2+3x+3)1/3>0 ∀x∈ℝ (2x2+3x+2)1/3>0 ∀x∈ℝ and 6x2+12x+8>0 ∀x∈ℝ then apply cauchy inequality (A+A1+A2+...+An)/n ≥(A*A1*A2*...*An)1/n; (A,A1,A2,...,An≥0) 6x2+12x+8=(1*1(x2+3x+3))1/3+(1*1(2x2+3x+2))1/3≤(x2+3x+3+1+1)/3+(2x2+3x+2+1+1)/3=x2+2x+3...

Use cauchy inequality, we can solve this easily.

solve(*) to find the final root (also = -1). to end the problem you must find all the roots or the equation becomes an impossible equation. and there is 1 root left in (*). In brief, i want you guys to help me solve (*) (normally do it like you didnt know about x=-1)

for A= (x2+3x+2)1/3 B= (2x2+3x+2)1/3 then thanks to calculator i know that x = -1 so i substitute x for -1 to find value of A,B then subtract exactly that value to find common factor (x+1): A -1 + B -1 = 6x2 +12x+6 (A-1)(A2+A*1+12)/(A2+A*1+12) + (B - 1)(B2+B*1+12)/(B2+B*1+12)-6(x+1)2=0...

Homework Statement (x2+3x+3)1/3 + (2x2+3x+2)1/3 = 6x2+12x+8 2.Relevant equationsThe Attempt at a Solution (x2+3x+3)1/3 -1 +(2x2+3x+2)1/3 -1 = 6x2+12x+6 (x2+3x+2)/((((x2+3x+3)1/3)2 + (x2+3x+3)1/3 +1) + (2x2 +3x+1)/((((2x2+3x+2)1/3)2)+(2x2+3x+2)1/3 +1) -6(x+1)2=0 then x=-1 or...