Precalculus Examples

$2 x^{2} + 3 x = 5$ , $(- 1, 5)$

Step 1

Subtract $5$ from both sides of the equation.

$2 x^{2} + 3 x - 5 = 0$

Step 2

Factor by grouping.

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Step 2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 3$ .

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Step 2.1.1

Factor $3$ out of $3 x$ .

$2 x^{2} + 3 (x) - 5 = 0$

Step 2.1.2

Rewrite $3$ as $- 2$ plus $5$

$2 x^{2} + (- 2 + 5) x - 5 = 0$

Step 2.1.3

Apply the distributive property.

$2 x^{2} - 2 x + 5 x - 5 = 0$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 2 x) + 5 x - 5 = 0$

Step 2.2.2

Factor out the greatest common factor (GCF) from each group.

$2 x (x - 1) + 5 (x - 1) = 0$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (2 x + 5) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$2 x + 5 = 0$

Step 4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $2 x + 5$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Step 5.2

Solve $2 x + 5 = 0$ for $x$ .

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Step 5.2.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 5.2.2

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Step 5.2.2.1

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 6

The final solution is all the values that make $(x - 1) (2 x + 5) = 0$ true.

$x = 1, - \frac{5}{2}$

Step 7

Find the values of $n$ that produce a value within the interval $(- 1, 5)$ .

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Step 7.1

The interval $(- 1, 5)$ does not contain $- \frac{5}{2}$ . It is not part of the final solution.

$- \frac{5}{2}$ is not on the interval

Step 7.2

The interval $(- 1, 5)$ contains $1$ .

$x = 1$

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