Find Where Undefined/Discontinuous y=1/(2x^2+3x-7)

$y = \frac{1}{2 x^{2} + 3 x - 7}$

Step 1

Set the denominator in $\frac{1}{2 x^{2} + 3 x - 7}$ equal to $0$ to find where the expression is undefined.

$2 x^{2} + 3 x - 7 = 0$

Step 2

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 $a = 2$ , $b = 3$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (2 \cdot - 7)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{- 3 \pm \sqrt{9 + 56}}{2 \cdot 2}$

Add $9$ and $56$ .

$x = \frac{- 3 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 3 \pm \sqrt{65}}{4}$

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

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{- 3 \pm \sqrt{9 + 56}}{2 \cdot 2}$

Add $9$ and $56$ .

$x = \frac{- 3 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 3 \pm \sqrt{65}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{- 3 + \sqrt{65}}{4}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 + \sqrt{65}}{4}$

Factor $- 1$ out of $\sqrt{65}$ .

$x = \frac{- 1 \cdot 3 - 1 (- \sqrt{65})}{4}$

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{65})$ .

$x = \frac{- 1 (3 - \sqrt{65})}{4}$

Move the negative in front of the fraction.

$x = - \frac{3 - \sqrt{65}}{4}$

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

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{- 3 \pm \sqrt{9 + 56}}{2 \cdot 2}$

Add $9$ and $56$ .

$x = \frac{- 3 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 3 \pm \sqrt{65}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{- 3 - \sqrt{65}}{4}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 - \sqrt{65}}{4}$

Factor $- 1$ out of $- \sqrt{65}$ .

$x = \frac{- 1 \cdot 3 - (\sqrt{65})}{4}$

Factor $- 1$ out of $- 1 (3) - (\sqrt{65})$ .

$x = \frac{- 1 (3 + \sqrt{65})}{4}$

Move the negative in front of the fraction.

$x = - \frac{3 + \sqrt{65}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{65}}{4}, - \frac{3 + \sqrt{65}}{4}$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - \frac{3 + \sqrt{65}}{4}, x = - \frac{3 - \sqrt{65}}{4}$

Step 4

Find Where Undefined/Discontinuous y=1/(2x^2+3x-7)

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