Solve 1+y+y-y^2

$$1 + y + y - y ^ { 2 }$$

Factor

$$-\left(y-\left(1-\sqrt{2}\right)\right)\left(y-\left(\sqrt{2}+1\right)\right)$$

Steps Using the Quadratic Formula

Combine y and y to get 2y.

$$factor(1+2y-y^{2})$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-y^{2}+2y+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-2±\sqrt{2^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square 2.

$$y=\frac{-2±\sqrt{4-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-2±\sqrt{4+4}}{2\left(-1\right)}$$

Add 4 to 4.

$$y=\frac{-2±\sqrt{8}}{2\left(-1\right)}$$

Take the square root of 8.

$$y=\frac{-2±2\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-2±2\sqrt{2}}{-2}$$

Now solve the equation y=\frac{-2±2\sqrt{2}}{-2} when ± is plus. Add -2 to 2\sqrt{2}.

$$y=\frac{2\sqrt{2}-2}{-2}$$

Divide -2+2\sqrt{2} by -2.

$$y=1-\sqrt{2}$$

Now solve the equation y=\frac{-2±2\sqrt{2}}{-2} when ± is minus. Subtract 2\sqrt{2} from -2.

$$y=\frac{-2\sqrt{2}-2}{-2}$$

Divide -2-2\sqrt{2} by -2.

$$y=\sqrt{2}+1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1-\sqrt{2} for x_{1} and 1+\sqrt{2} for x_{2}.

$$-y^{2}+2y+1=-\left(y-\left(1-\sqrt{2}\right)\right)\left(y-\left(\sqrt{2}+1\right)\right)$$

Evaluate

$$1+2y-y^{2}$$

Solution Steps

Combine y and y to get 2y.

$$1+2y-y^{2}$$