Find the Points of Intersection y=x^2+5x-2 , y=x+1

$y = x^{2} + 5 x - 2$ , $y = x + 1$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} + 5 x - 2 = x + 1$

Step 2

Solve $x^{2} + 5 x - 2 = x + 1$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$x^{2} + 5 x - 2 - x = 1$

Subtract $x$ from $5 x$ .

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

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

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $- 2$ .

$x^{2} + 4 x - 3 = 0$

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 $a = 1$ , $b = 4$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 3)}}{2 \cdot 1} + y = x + 1$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 1}$

Add $16$ and $12$ .

$x = \frac{- 4 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{2}$ .

$x = - 2 \pm \sqrt{7}$

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

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 1}$

Add $16$ and $12$ .

$x = \frac{- 4 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{2}$ .

$x = - 2 \pm \sqrt{7}$

Change the $\pm$ to $+$ .

$x = - 2 + \sqrt{7}$

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

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 1}$

Add $16$ and $12$ .

$x = \frac{- 4 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{2}$ .

$x = - 2 \pm \sqrt{7}$

Change the $\pm$ to $-$ .

$x = - 2 - \sqrt{7}$

The final answer is the combination of both solutions.

$x = - 2 + \sqrt{7}, - 2 - \sqrt{7}$

Step 3

Evaluate $y$ when $x = - 2 + \sqrt{7}$ .

Substitute $- 2 + \sqrt{7}$ for $x$ .

$y = (- 2 + \sqrt{7}) + 1$

Substitute $- 2 + \sqrt{7}$ for $x$ in $y = (- 2 + \sqrt{7}) + 1$ and solve for $y$ .

Remove parentheses.

$y = - 2 + \sqrt{7} + 1$

Remove parentheses.

$y = (- 2 + \sqrt{7}) + 1$

Add $- 2$ and $1$ .

$y = - 1 + \sqrt{7}$

Step 4

Evaluate $y$ when $x = - 2 - \sqrt{7}$ .

Substitute $- 2 - \sqrt{7}$ for $x$ .

$y = (- 2 - \sqrt{7}) + 1$

Substitute $- 2 - \sqrt{7}$ for $x$ in $y = (- 2 - \sqrt{7}) + 1$ and solve for $y$ .

Remove parentheses.

$y = - 2 - \sqrt{7} + 1$

Remove parentheses.

$y = (- 2 - \sqrt{7}) + 1$

Add $- 2$ and $1$ .

$y = - 1 - \sqrt{7}$

Step 5

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

$(- 2 + \sqrt{7}, - 1 + \sqrt{7})$

$(- 2 - \sqrt{7}, - 1 - \sqrt{7})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 2 + \sqrt{7}, - 1 + \sqrt{7}), (- 2 - \sqrt{7}, - 1 - \sqrt{7})$

Equation Form:

$x = - 2 + \sqrt{7}, y = - 1 + \sqrt{7}$

$x = - 2 - \sqrt{7}, y = - 1 - \sqrt{7}$

Step 7

Find the Points of Intersection y=x^2+5x-2 , y=x+1

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