Solve 1+a^2=(sqrt{2}a-sqrt{2})^2

$$1 + a ^ { 2 } = ( \sqrt { 2 } a - \sqrt { 2 } ) ^ { 2 }$$

Solve for a

$$a=\sqrt{3}+2\approx 3.732050808$$

$$a=2-\sqrt{3}\approx 0.267949192$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}a-\sqrt{2}\right)^{2}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

The square of \sqrt{2} is 2.

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

Multiply -2 and 2 to get -4.

$$1+a^{2}=2a^{2}-4a+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$1+a^{2}=2a^{2}-4a+2$$

Subtract 2a^{2} from both sides.

$$1+a^{2}-2a^{2}=-4a+2$$

Combine a^{2} and -2a^{2} to get -a^{2}.

$$1-a^{2}=-4a+2$$

Add 4a to both sides.

$$1-a^{2}+4a=2$$

Subtract 2 from both sides.

$$1-a^{2}+4a-2=0$$

Subtract 2 from 1 to get -1.

$$-1-a^{2}+4a=0$$

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.

$$-a^{2}+4a-1=0$$

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

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

Square 4.

$$a=\frac{-4±\sqrt{16-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

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

Multiply 4 times -1.

$$a=\frac{-4±\sqrt{16-4}}{2\left(-1\right)}$$

Add 16 to -4.

$$a=\frac{-4±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

$$a=\frac{-4±2\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-4±2\sqrt{3}}{-2}$$

Now solve the equation a=\frac{-4±2\sqrt{3}}{-2} when ± is plus. Add -4 to 2\sqrt{3}.

$$a=\frac{2\sqrt{3}-4}{-2}$$

Divide -4+2\sqrt{3} by -2.

$$a=2-\sqrt{3}$$

Now solve the equation a=\frac{-4±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from -4.

$$a=\frac{-2\sqrt{3}-4}{-2}$$

Divide -4-2\sqrt{3} by -2.

$$a=\sqrt{3}+2$$

The equation is now solved.

$$a=2-\sqrt{3} a=\sqrt{3}+2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}a-\sqrt{2}\right)^{2}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

The square of \sqrt{2} is 2.

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

Multiply -2 and 2 to get -4.

$$1+a^{2}=2a^{2}-4a+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$1+a^{2}=2a^{2}-4a+2$$

Subtract 2a^{2} from both sides.

$$1+a^{2}-2a^{2}=-4a+2$$

Combine a^{2} and -2a^{2} to get -a^{2}.

$$1-a^{2}=-4a+2$$

Add 4a to both sides.

$$1-a^{2}+4a=2$$

Subtract 1 from both sides.

$$-a^{2}+4a=2-1$$

Subtract 1 from 2 to get 1.

$$-a^{2}+4a=1$$

Divide both sides by -1.

$$\frac{-a^{2}+4a}{-1}=\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\frac{4}{-1}a=\frac{1}{-1}$$

Divide 4 by -1.

$$a^{2}-4a=\frac{1}{-1}$$

Divide 1 by -1.

$$a^{2}-4a=-1$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-4a+\left(-2\right)^{2}=-1+\left(-2\right)^{2}$$

Square -2.

$$a^{2}-4a+4=-1+4$$

Add -1 to 4.

$$a^{2}-4a+4=3$$

Factor a^{2}-4a+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-2\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{3}$$

Simplify.

$$a-2=\sqrt{3} a-2=-\sqrt{3}$$

Add 2 to both sides of the equation.

$$a=\sqrt{3}+2 a=2-\sqrt{3}$$