Find the Points of Intersection y=x^2-3 , y=x+1

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

Eliminate the equal sides of each equation and combine.

$x^{2} - 3 = x + 1$

Solve $x^{2} - 3 = x + 1$ for $x$ .

Subtract $x$ from both sides of the equation.

$x^{2} - 3 - x = 1$

Move all terms to the left side of the equation and simplify.

Subtract $1$ from both sides of the equation.

$x^{2} - 3 - x - 1 = 0$

Subtract $1$ from $- 3$ .

$x^{2} - x - 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 1$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{1 \pm \sqrt{1 + 16}}{2 \cdot 1}$

Add $1$ and $16$ .

$x = \frac{1 \pm \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{17}}{2}$

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

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{1 \pm \sqrt{1 + 16}}{2 \cdot 1}$

Add $1$ and $16$ .

$x = \frac{1 \pm \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{17}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{1 + \sqrt{17}}{2}$

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

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{1 \pm \sqrt{1 + 16}}{2 \cdot 1}$

Add $1$ and $16$ .

$x = \frac{1 \pm \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{17}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{1 - \sqrt{17}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{17}}{2}, \frac{1 - \sqrt{17}}{2}$

Evaluate $y$ when $x = \frac{1 + \sqrt{17}}{2}$ .

Substitute $\frac{1 + \sqrt{17}}{2}$ for $x$ .

$y = (\frac{1 + \sqrt{17}}{2}) + 1$

Substitute $\frac{1 + \sqrt{17}}{2}$ for $x$ in $y = (\frac{1 + \sqrt{17}}{2}) + 1$ and solve for $y$ .

Remove parentheses.

$y = \frac{1 + \sqrt{17}}{2} + 1$

Remove parentheses.

$y = (\frac{1 + \sqrt{17}}{2}) + 1$

Simplify $(\frac{1 + \sqrt{17}}{2}) + 1$ .

Write $1$ as a fraction with a common denominator.

$y = \frac{1 + \sqrt{17}}{2} + \frac{2}{2}$

Combine the numerators over the common denominator.

$y = \frac{1 + \sqrt{17} + 2}{2}$

Add $1$ and $2$ .

$y = \frac{3 + \sqrt{17}}{2}$

Evaluate $y$ when $x = \frac{1 - \sqrt{17}}{2}$ .

Substitute $\frac{1 - \sqrt{17}}{2}$ for $x$ .

$y = (\frac{1 - \sqrt{17}}{2}) + 1$

Substitute $\frac{1 - \sqrt{17}}{2}$ for $x$ in $y = (\frac{1 - \sqrt{17}}{2}) + 1$ and solve for $y$ .

Remove parentheses.

$y = \frac{1 - \sqrt{17}}{2} + 1$

Remove parentheses.

$y = (\frac{1 - \sqrt{17}}{2}) + 1$

Simplify $(\frac{1 - \sqrt{17}}{2}) + 1$ .

Write $1$ as a fraction with a common denominator.

$y = \frac{1 - \sqrt{17}}{2} + \frac{2}{2}$

Combine the numerators over the common denominator.

$y = \frac{1 - \sqrt{17} + 2}{2}$

Add $1$ and $2$ .

$y = \frac{3 - \sqrt{17}}{2}$

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

$(\frac{1 + \sqrt{17}}{2}, \frac{3 + \sqrt{17}}{2})$

$(\frac{1 - \sqrt{17}}{2}, \frac{3 - \sqrt{17}}{2})$

The result can be shown in multiple forms.

Point Form:

$(\frac{1 + \sqrt{17}}{2}, \frac{3 + \sqrt{17}}{2}), (\frac{1 - \sqrt{17}}{2}, \frac{3 - \sqrt{17}}{2})$

Equation Form:

$x = \frac{1 + \sqrt{17}}{2}, y = \frac{3 + \sqrt{17}}{2}$

$x = \frac{1 - \sqrt{17}}{2}, y = \frac{3 - \sqrt{17}}{2}$

Find the Points of Intersection y=x^2-3 , y=x+1

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