x*(x+1) - 2*(x+1) = 4
Solve x.
\(\begin{array}{rcl} x\cdot (x+1) - 2\cdot (x+1) &=& 4 \\ x^2+x-2x-2 &=& 4\\ x^2-x-6 &=& 0\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a}~ }\\ x^2-x-6 &=& 0\\ x &=& {1 \pm \sqrt{1-4(-6)} \over 2} \\ x &=& {1 \pm \sqrt{1+24} \over 2} \\ x &=& {1 \pm \sqrt{25} \over 2} \\ x &=& {1 \pm 5 \over 2} \\ \hline x_1 &=& {1 + 5 \over 2}\\ x_1 &=& {6 \over 2}\\ \mathbf{x_1} &\mathbf{=} & \mathbf{3}\\ \hline x_2 &=& {1 - 5 \over 2}\\ x_2 &=& {-4 \over 2}\\ \mathbf{x_2} &\mathbf{=} & \mathbf{-2}\\ \end{array}\)
Solve for x:
x (x+1)-2 (x+1) = 4
Expand out terms of the left hand side:
x^2-x-2 = 4
Add 2 to both sides:
x^2-x = 6
Add 1/4 to both sides:
x^2-x+1/4 = 25/4
Write the left hand side as a square:
(x-1/2)^2 = 25/4
Take the square root of both sides:
x-1/2 = 5/2 or x-1/2 = -5/2
Add 1/2 to both sides:
x = 3 or x-1/2 = -5/2
Add 1/2 to both sides:
Answer: |
| x = 3 or x = -2
x*(x+1) - 2*(x+1) = 4
Solve x.
\(\begin{array}{rcl} x\cdot (x+1) - 2\cdot (x+1) &=& 4 \\ x^2+x-2x-2 &=& 4\\ x^2-x-6 &=& 0\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a}~ }\\ x^2-x-6 &=& 0\\ x &=& {1 \pm \sqrt{1-4(-6)} \over 2} \\ x &=& {1 \pm \sqrt{1+24} \over 2} \\ x &=& {1 \pm \sqrt{25} \over 2} \\ x &=& {1 \pm 5 \over 2} \\ \hline x_1 &=& {1 + 5 \over 2}\\ x_1 &=& {6 \over 2}\\ \mathbf{x_1} &\mathbf{=} & \mathbf{3}\\ \hline x_2 &=& {1 - 5 \over 2}\\ x_2 &=& {-4 \over 2}\\ \mathbf{x_2} &\mathbf{=} & \mathbf{-2}\\ \end{array}\)