Find the Roots (Zeros) y=x^2-2 square root of 2x+2

$y = x^{2} - 2 \sqrt{2} x + 2$

Set $x^{2} - 2 \sqrt{2} x + 2$ equal to $0$ .

$x^{2} - 2 \sqrt{2} x + 2 = 0$

Solve for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2 \sqrt{2}$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{2 \sqrt{2} \pm \sqrt{{(- 2 \sqrt{2})}^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Apply the product rule to $- 2 \sqrt{2}$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{{(- 2)}^{2} {\sqrt{2}}^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{4 {\sqrt{2}}^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{4 {(2^{\frac{1}{2}})}^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{4 \cdot 2^{\frac{1}{2} \cdot 2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Combine $\frac{1}{2}$ and $2$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{4 \cdot 2^{\frac{2}{2}} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \frac{2 \sqrt{2} \pm \sqrt{4 \cdot 2^{\frac{2}{2}} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Rewrite the expression.

$x = \frac{2 \sqrt{2} \pm \sqrt{4 \cdot 2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Evaluate the exponent.

$x = \frac{2 \sqrt{2} \pm \sqrt{4 \cdot 2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $4$ by $2$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{8 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{8 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{8 - 8}}{2 \cdot 1}$

Subtract $8$ from $8$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{2 \sqrt{2} \pm \sqrt{0^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \sqrt{2} \pm 0}{2 \cdot 1}$

$2 \sqrt{2}$ plus or minus $0$ is $2 \sqrt{2}$ .

$x = \frac{2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \sqrt{2}}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \frac{2 \sqrt{2}}{2}$

Divide $\sqrt{2}$ by $1$ .

$x = \sqrt{2}$

The final answer is the combination of both solutions.

$x = \sqrt{2}$ Double roots

Find the Roots (Zeros) y=x^2-2 square root of 2x+2

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