Question

Solve this equation algebraically. (x+1)^2 -2=2/x

Answer 1

x = -1, 1 and -2.

Explanation:

(x+1)^2 - 2 = 2/x

Multiply through by x:

x(x + 1)^2 - 2x = 2

x(x^2 + 2x + 1) - 2x - 2 = 0

x^3 + 2x^2 + x - 2x - 2 = 0

x^3 + 2x^2 - x - 2 = 0

x^2(x + 2) - 1(x + 2) = 0

(x^2 - 1)(x + 2) = 0

(x + 1)(x - 1)(x + 2) = 0

These gives x = -1, 1 and -2.

Answer 2

`\huge\boxed{x=-2\ \vee\ x=-1\ \vee\ x=1}`

Explanation:

`(x+1)^2-2=(2)/(x)\\\\\text{Domain:}\ x\\eq0\\\\((x+1)^2-2)/(1)=(2)/(x)\qquad\text{cross multiply}\\\\x\bigg((x+1)^2-2\bigg)=(1)(2)\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\x\bigg(x^2+(2)(x)(1)+1^2-2\bigg)=2\\\\x\bigg(x^2+2x+1-2\bigg)=2\\\\x\bigg(x^2+2x-1\bigg)=2\qquad\text{use the distributive property}\\\\(x)(x^2)+(x)(2x)+(x)(-1)=2`

`x^3+2x^2-x=2\qquad\text{subtract 2 from both sides}\\\\x^3+2x^2-x-2=0\qquad\text{distributive}\\\\x^2(x+2)-1(x+2)=0\qquad\text{distributive}\\\\(x+2)(x^2-1)=0`

`\text{The product is equal 0 if one of factors is equal 0.}\\\text{Therefore}\\\\(x+2)(x^2-1)=0\iff x+2=0\ \vee\ x^2-1=0\\\\x+2=0\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2}\in\text{Domain}\\\\x^2-1=0\qquad\text{add 1 to both sides}\\\\x^2=1\Rightarrow x=\pm\sqrt1\\\\\boxed{x=-1}\in\text{Domain}\ \vee\ \boxed{x=1}\in\text{Domain}`