Question

Solve for x.

$\frac{2(x-1)+2(3x^2-4x)(x^3-2x^2)}{2\sqrt{(x-1)^2+(x^3-2x^2)^2}}=0$

Answer 1

Step-by-step explanation:

$\frac{2(x-1)+2(3x^2-4x)(x^3-2x^2)}{2\sqrt{(x-1)^2+(x^3-2x^2)^2}}=0$

$=>2(x-1)+2(3x^2-4x)(x^3-2x^2)=0\\=>(x-1)+(3x^2-4x)(x^3-2x^2)=0\\=>(x+1)+(3x^5-6x^4-4x^4+8x^3)=0\\=>(x+1)+(3x^5-10x^4+8x^3)=0\\=>(x+1)+(3x^5-6x4-4x^4+8x^3)=0\\=>(x+1)+(3x^4(x-2)-4x^3(x-2)=0\\=>(x+1)+(x-2)(3x^4-4x^3)=0\\=>(x+1)+(x-2)x^3(3x-4)=0\\=>(x+1)+x^3(x-2)(3x-4)=0$

Answer 2

Answer:

$\frac{2(x-1)+2(3x^2-4x)(x^3-2x^2)}{2\sqrt{(x-1)^2+(x^3-2x^2)^2}}=0$

➡️$\green{2(x-1)+2(3x^2-4x)(x^3-2x^2)}=0$

➡️$\pink{2x-2+(6x^2-8x)(x^3-2x^2)=0}$

➡️$\blue{2x-2+6x^5-12x^4-8x^4+16x^3} =0$

➡️$\red{6x^5-20x^4+16x^3+2x-2}=0$

➡️$\green{2(3x^5-10x^4+8x^3+x-1)}=0$

➡️$\orange{(x-1)+(3x^5-10x^4+8x^3)}=0$

➡️$\blue{(x-1)+(3x^5-6x^4-4x^4+8x^3)}=0$

➡️$\pink{(x-1)+(3x^4(x-2)-4x^3(x-2)}=0$

➡️$\green{(x-1)+(x-2)(3x^4-4x^3)}=0$

➡️$\red{(x-1)+(x-2)x^3(3x-4)}=0$

➡️$\blue{3x^5-10x^4+8x^3+x-1)}=0$

$\orange{So,\: the\: x \:values \:are \:0.51 (or)\\ 1.41\:(or)\: 1.92\: (or) \:-0.26+-0.41i}$