Solve {l}{y=k(x+1)}{x^2/2+y^2=1}

$$\left\{ \begin{array} { l } { y = k ( x + 1 ) } \\ { \frac { x ^ { 2 } } { 2 } + y ^ { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{k^{2}+1}\right)}{2k^{2}+1}\text{, }y=\frac{k\left(-\sqrt{2\left(k^{2}+1\right)}+1\right)}{2k^{2}+1}$$

$$x=\frac{-2k^{2}+\sqrt{2\left(k^{2}+1\right)}}{2k^{2}+1}\text{, }y=\frac{k\left(\sqrt{2\left(k^{2}+1\right)}+1\right)}{2k^{2}+1}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Use the distributive property to multiply k by x+1.

$$y=kx+k$$

Substitute kx+k for y in the other equation, x^{2}+2y^{2}=2.

$$x^{2}+2\left(kx+k\right)^{2}=2$$

Square kx+k.

$$x^{2}+2\left(k^{2}x^{2}+2kkx+k^{2}\right)=2$$

Multiply 2 times k^{2}x^{2}+2kkx+k^{2}.

$$x^{2}+2k^{2}x^{2}+4k^{2}x+2k^{2}=2$$

Add x^{2} to 2k^{2}x^{2}.

$$\left(2k^{2}+1\right)x^{2}+4k^{2}x+2k^{2}=2$$

Subtract 2 from both sides of the equation.

$$\left(2k^{2}+1\right)x^{2}+4k^{2}x+2k^{2}-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+2k^{2} for a, 2\times 2kk for b, and 2k^{2}-2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2\times 2kk.

$$x=\frac{-4k^{2}±\sqrt{16k^{4}-4\left(2k^{2}+1\right)\left(2k^{2}-2\right)}}{2\left(2k^{2}+1\right)}$$

Multiply -4 times 1+2k^{2}.

$$x=\frac{-4k^{2}±\sqrt{16k^{4}+\left(-8k^{2}-4\right)\left(2k^{2}-2\right)}}{2\left(2k^{2}+1\right)}$$

Multiply -4-8k^{2} times 2k^{2}-2.

$$x=\frac{-4k^{2}±\sqrt{16k^{4}+8+8k^{2}-16k^{4}}}{2\left(2k^{2}+1\right)}$$

Add 16k^{4} to -16k^{4}+8k^{2}+8.

$$x=\frac{-4k^{2}±\sqrt{8k^{2}+8}}{2\left(2k^{2}+1\right)}$$

Take the square root of 8k^{2}+8.

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

Multiply 2 times 1+2k^{2}.

$$x=\frac{-4k^{2}±2\sqrt{2k^{2}+2}}{4k^{2}+2}$$

Now solve the equation x=\frac{-4k^{2}±2\sqrt{2k^{2}+2}}{4k^{2}+2} when ± is plus. Add -4k^{2} to 2\sqrt{2k^{2}+2}.

$$x=\frac{-4k^{2}+2\sqrt{2k^{2}+2}}{4k^{2}+2}$$

Divide -4k^{2}+2\sqrt{2k^{2}+2} by 2+4k^{2}.

$$x=\frac{-2k^{2}+\sqrt{2k^{2}+2}}{2k^{2}+1}$$

Now solve the equation x=\frac{-4k^{2}±2\sqrt{2k^{2}+2}}{4k^{2}+2} when ± is minus. Subtract 2\sqrt{2k^{2}+2} from -4k^{2}.

$$x=\frac{-4k^{2}-2\sqrt{2k^{2}+2}}{4k^{2}+2}$$

Divide -4k^{2}-2\sqrt{2k^{2}+2} by 2+4k^{2}.

$$x=-\frac{2k^{2}+\sqrt{2k^{2}+2}}{2k^{2}+1}$$

There are two solutions for x: \frac{-2k^{2}+\sqrt{2+2k^{2}}}{1+2k^{2}} and -\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{1+k^{2}}\right)}{1+2k^{2}}. Substitute \frac{-2k^{2}+\sqrt{2+2k^{2}}}{1+2k^{2}} for x in the equation y=kx+k to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{-2k^{2}+\sqrt{2k^{2}+2}}{2k^{2}+1}+k$$

Multiply k times \frac{-2k^{2}+\sqrt{2+2k^{2}}}{1+2k^{2}}.

$$y=\frac{-2k^{2}+\sqrt{2k^{2}+2}}{2k^{2}+1}k+k$$

Now substitute -\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{1+k^{2}}\right)}{1+2k^{2}} for x in the equation y=kx+k and solve to find the corresponding solution for y that satisfies both equations.

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

Multiply k times -\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{1+k^{2}}\right)}{1+2k^{2}}.

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

The system is now solved.

$$y=\frac{-2k^{2}+\sqrt{2k^{2}+2}}{2k^{2}+1}k+k,x=\frac{-2k^{2}+\sqrt{2k^{2}+2}}{2k^{2}+1}\text{ or }y=\left(-\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{k^{2}+1}\right)}{2k^{2}+1}\right)k+k,x=-\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{k^{2}+1}\right)}{2k^{2}+1}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{k^{2}+1}\right)}{2k^{2}+1}\text{, }y=\frac{k\left(-\sqrt{2\left(k^{2}+1\right)}+1\right)}{2k^{2}+1}\text{; }x=\frac{-2k^{2}+\sqrt{2\left(k^{2}+1\right)}}{2k^{2}+1}\text{, }y=\frac{k\left(\sqrt{2\left(k^{2}+1\right)}+1\right)}{2k^{2}+1}\text{, }&k\neq -\frac{\sqrt{2}i}{2}\text{ and }k\neq \frac{\sqrt{2}i}{2}\\x=\frac{1-k^{2}}{2k^{2}}\text{, }y=\frac{k^{2}+1}{2k}\text{, }&k=-\frac{\sqrt{2}i}{2}\text{ or }k=\frac{\sqrt{2}i}{2}\end{matrix}\right.$$