Solve the System of Equations y=x^2 y=-2x+1

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

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} = - 2 x + 1$

Step 2

Solve $x^{2} = - 2 x + 1$ for $x$ .

Add $2 x$ to both sides of the equation.

$x^{2} + 2 x = 1$

Subtract $1$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a} + y = - 2 x + 1$

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $4$ and $4$ .

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

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $4$ and $4$ .

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

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Change the $\pm$ to $+$ .

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

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $4$ and $4$ .

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

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Step 3

Evaluate $y$ when $x = - 1 + \sqrt{2}$ .

Substitute $- 1 + \sqrt{2}$ for $x$ .

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

Simplify $- 2 (- 1 + \sqrt{2}) + 1$ .

Simplify each term.

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Add $2$ and $1$ .

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

Step 4

Evaluate $y$ when $x = - 1 - \sqrt{2}$ .

Substitute $- 1 - \sqrt{2}$ for $x$ .

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

Simplify $- 2 (- 1 - \sqrt{2}) + 1$ .

Simplify each term.

Apply the distributive property.

$y = - 2 \cdot - 1 - 2 (- \sqrt{2}) + 1$

Multiply $- 2$ by $- 1$ .

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

Multiply $- 1$ by $- 2$ .

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

Add $2$ and $1$ .

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

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 1 + \sqrt{2}, 3 - 2 \sqrt{2})$

$(- 1 - \sqrt{2}, 3 + 2 \sqrt{2})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 1 + \sqrt{2}, 3 - 2 \sqrt{2}), (- 1 - \sqrt{2}, 3 + 2 \sqrt{2})$

Equation Form:

$x = - 1 + \sqrt{2}, y = 3 - 2 \sqrt{2}$

$x = - 1 - \sqrt{2}, y = 3 + 2 \sqrt{2}$

Step 7

Solve the System of Equations y=x^2 y=-2x+1

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