Precalculus Examples

x^{2} + 2 x - 1

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

Step 1

Find the discriminant for

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

$x^{2} + 2 x - 1 = 0$ . In this case,

b^{2} - 4 a c = 8

$b^{2} - 4 a c = 8$ .

Tap for more steps...

Step 1.1

The discriminant of a quadratic is the expression inside the radical of the quadratic formula.

b^{2} - 4 (a c)

$b^{2} - 4 (a c)$

Step 1.2

Substitute in the values of

a

$a$ ,

b

$b$ , and

c

$c$ .

2^{2} - 4 (1 \cdot - 1)

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

Step 1.3

Evaluate the result to find the discriminant.

Tap for more steps...

Step 1.3.1

Simplify each term.

Tap for more steps...

Step 1.3.1.1

Raise

2

$2$ to the power of

2

$2$ .

4 - 4 (1 \cdot - 1)

$4 - 4 (1 \cdot - 1)$

Step 1.3.1.2

Multiply

- 4 (1 \cdot - 1)

$- 4 (1 \cdot - 1)$ .

Tap for more steps...

Step 1.3.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

4 - 4 \cdot - 1

$4 - 4 \cdot - 1$

Step 1.3.1.2.2

Multiply

- 4

$- 4$ by

- 1

$- 1$ .

4 + 4

$4 + 4$

4 + 4

$4 + 4$

4 + 4

$4 + 4$

Step 1.3.2

Add

4

$4$ and

4

$4$ .

8

$8$

8

$8$

8

$8$

Step 2

A perfect square number is an integer that is the square of another integer.

\sqrt{8} \approx 2.82842712

$\sqrt{8} \approx 2.82842712$ , which is not an integer number.

\sqrt{8} \approx 2.82842712

$\sqrt{8} \approx 2.82842712$

Step 3

Since

8

$8$ can't be the square of another integer, it is not a perfect square number.

Step 4

The polynomial

x^{2} + 2 x - 1

$x^{2} + 2 x - 1$ is prime because the discriminant is not a perfect square number.

Prime