Solve (sqrt{2}x-3)(sqrt{2}x+3)=2x

$$( \sqrt { 2 } x - 3 ) ( \sqrt { 2 } x + 3 ) = 2 x$$

Solve for x

$$x = \frac{\sqrt{19} + 1}{2} \approx 2.679449472$$

$$x=\frac{1-\sqrt{19}}{2}\approx -1.679449472$$

Steps Using the Quadratic Formula

Consider \left(\sqrt{2}x-3\right)\left(\sqrt{2}x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(\sqrt{2}x\right)^{2}-9=2x$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\left(\sqrt{2}\right)^{2}x^{2}-9=2x$$

The square of \sqrt{2} is 2.

$$2x^{2}-9=2x$$

Subtract 2x from both sides.

$$2x^{2}-9-2x=0$$

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.

$$2x^{2}-2x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-9\right)}}{2\times 2}$$

Multiply -8 times -9.

$$x=\frac{-\left(-2\right)±\sqrt{4+72}}{2\times 2}$$

Add 4 to 72.

$$x=\frac{-\left(-2\right)±\sqrt{76}}{2\times 2}$$

Take the square root of 76.

$$x=\frac{-\left(-2\right)±2\sqrt{19}}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{19}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±2\sqrt{19}}{4}$$

Now solve the equation x=\frac{2±2\sqrt{19}}{4} when ± is plus. Add 2 to 2\sqrt{19}.

$$x=\frac{2\sqrt{19}+2}{4}$$

Divide 2+2\sqrt{19} by 4.

$$x=\frac{\sqrt{19}+1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{19}}{4} when ± is minus. Subtract 2\sqrt{19} from 2.

$$x=\frac{2-2\sqrt{19}}{4}$$

Divide 2-2\sqrt{19} by 4.

$$x=\frac{1-\sqrt{19}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{19}+1}{2} x=\frac{1-\sqrt{19}}{2}$$

Steps for Completing the Square

Consider \left(\sqrt{2}x-3\right)\left(\sqrt{2}x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(\sqrt{2}x\right)^{2}-9=2x$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\left(\sqrt{2}\right)^{2}x^{2}-9=2x$$

The square of \sqrt{2} is 2.

$$2x^{2}-9=2x$$

Subtract 2x from both sides.

$$2x^{2}-9-2x=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$2x^{2}-2x=9$$

Divide both sides by 2.

$$\frac{2x^{2}-2x}{2}=\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{2}{2}\right)x=\frac{9}{2}$$

Divide -2 by 2.

$$x^{2}-x=\frac{9}{2}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{9}{2}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{9}{2}+\frac{1}{4}$$

Add \frac{9}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{19}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{19}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{19}}{2} x-\frac{1}{2}=-\frac{\sqrt{19}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{19}+1}{2} x=\frac{1-\sqrt{19}}{2}$$