Algebra Examples

x^{2} - 49

$x^{2} - 49$

Step 1

Find the discriminant for

x^{2} - 49 = 0

$x^{2} - 49 = 0$ . In this case,

b^{2} - 4 a c = 196

$b^{2} - 4 a c = 196$ .

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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

$c$ .

0^{2} - 4 (1 \cdot - 49)

$0^{2} - 4 (1 \cdot - 49)$

Step 1.3

Evaluate the result to find the discriminant.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Raising

0

$0$ to any positive power yields

0

$0$ .

0 - 4 (1 \cdot - 49)

$0 - 4 (1 \cdot - 49)$

Step 1.3.1.2

Multiply

- 4 (1 \cdot - 49)

$- 4 (1 \cdot - 49)$ .

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Step 1.3.1.2.1

Multiply

- 49

$- 49$ by

1

$1$ .

0 - 4 \cdot - 49

$0 - 4 \cdot - 49$

Step 1.3.1.2.2

Multiply

- 4

$- 4$ by

- 49

$- 49$ .

0 + 196

$0 + 196$

0 + 196

$0 + 196$

0 + 196

$0 + 196$

Step 1.3.2

Add

0

$0$ and

196

$196$ .

196

$196$

196

$196$

196

$196$

Step 2

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

\sqrt{196} = 14

$\sqrt{196} = 14$ , which is an integer number.

\sqrt{196} = 14

$\sqrt{196} = 14$

Step 3

Since

196

$196$ is the square of

14

$14$ , it is a perfect square number.

196

$196$ is a perfect square number

Step 4

The polynomial

x^{2} - 49

$x^{2} - 49$ is not prime because the discriminant is a perfect square number.

Not prime

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