Precalculus Examples

$z^{2} - 2 z - 24 = 0$

Step 1

Add $24$ to both sides of the equation.

$z^{2} - 2 z = 24$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- 1)}^{2}$

Step 3

Add the term to each side of the equation.

$z^{2} - 2 z + {(- 1)}^{2} = 24 + {(- 1)}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

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

$z^{2} - 2 z + 1 = 24 + {(- 1)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $24 + {(- 1)}^{2}$ .

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Step 4.2.1.1

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

$z^{2} - 2 z + 1 = 24 + 1$

Step 4.2.1.2

Add $24$ and $1$ .

$z^{2} - 2 z + 1 = 25$

Step 5

Factor the perfect trinomial square into ${(z - 1)}^{2}$ .

${(z - 1)}^{2} = 25$

Step 6

Solve the equation for $z$ .

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Step 6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$z - 1 = \pm \sqrt{25}$

Step 6.2

Simplify $\pm \sqrt{25}$ .

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Step 6.2.1

Rewrite $25$ as $5^{2}$ .

$z - 1 = \pm \sqrt{5^{2}}$

Step 6.2.2

Pull terms out from under the radical, assuming positive real numbers.

$z - 1 = \pm 5$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.3.1

First, use the positive value of the $\pm$ to find the first solution.

$z - 1 = 5$

Step 6.3.2

Move all terms not containing $z$ to the right side of the equation.

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Step 6.3.2.1

Add $1$ to both sides of the equation.

$z = 5 + 1$

Step 6.3.2.2

Add $5$ and $1$ .

$z = 6$

Step 6.3.3

Next, use the negative value of the $\pm$ to find the second solution.

$z - 1 = - 5$

Step 6.3.4

Move all terms not containing $z$ to the right side of the equation.

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Step 6.3.4.1

Add $1$ to both sides of the equation.

$z = - 5 + 1$

Step 6.3.4.2

Add $- 5$ and $1$ .

$z = - 4$

Step 6.3.5

The complete solution is the result of both the positive and negative portions of the solution.

$z = 6, - 4$