Basic Math Examples

∣ ∣ 2 d^{2} - 3 d ∣ ∣ = 5

$| 2 d^{2} - 3 d | = 5$

Step 1

Remove the absolute value term. This creates a

\pm

$\pm$ on the right side of the equation because

| x | = \pm x

$| x | = \pm x$ .

2 d^{2} - 3 d = \pm 5

$2 d^{2} - 3 d = \pm 5$

Step 2

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

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Step 2.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

2 d^{2} - 3 d = 5

$2 d^{2} - 3 d = 5$

Step 2.2

Subtract

5

$5$ from both sides of the equation.

2 d^{2} - 3 d - 5 = 0

$2 d^{2} - 3 d - 5 = 0$

Step 2.3

Factor by grouping.

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Step 2.3.1

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 2 \cdot - 5 = - 10

$a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is

b = - 3

$b = - 3$ .

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Step 2.3.1.1

Factor

- 3

$- 3$ out of

- 3 d

$- 3 d$ .

2 d^{2} - 3 d - 5 = 0

$2 d^{2} - 3 d - 5 = 0$

Step 2.3.1.2

Rewrite

- 3

$- 3$ as

2

$2$ plus

- 5

$- 5$

2 d^{2} + (2 - 5) d - 5 = 0

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

Step 2.3.1.3

Apply the distributive property.

2 d^{2} + 2 d - 5 d - 5 = 0

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

2 d^{2} + 2 d - 5 d - 5 = 0

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

Step 2.3.2

Factor out the greatest common factor from each group.

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Step 2.3.2.1

Group the first two terms and the last two terms.

(2 d^{2} + 2 d) - 5 d - 5 = 0

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

Step 2.3.2.2

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

2 d (d + 1) - 5 (d + 1) = 0

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

2 d (d + 1) - 5 (d + 1) = 0

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

Step 2.3.3

Factor the polynomial by factoring out the greatest common factor,

d + 1

$d + 1$ .

(d + 1) (2 d - 5) = 0

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

(d + 1) (2 d - 5) = 0

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

Step 2.4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

d + 1 = 0

$d + 1 = 0$

2 d - 5 = 0

$2 d - 5 = 0$

Step 2.5

Set

d + 1

$d + 1$ equal to

0

$0$ and solve for

d

$d$ .

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Step 2.5.1

Set

d + 1

$d + 1$ equal to

0

$0$ .

d + 1 = 0

$d + 1 = 0$

Step 2.5.2

Subtract

1

$1$ from both sides of the equation.

d = - 1

$d = - 1$

d = - 1

$d = - 1$

Step 2.6

Set

2 d - 5

$2 d - 5$ equal to

0

$0$ and solve for

d

$d$ .

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Step 2.6.1

Set

2 d - 5

$2 d - 5$ equal to

0

$0$ .

2 d - 5 = 0

$2 d - 5 = 0$

Step 2.6.2

Solve

2 d - 5 = 0

$2 d - 5 = 0$ for

d

$d$ .

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Step 2.6.2.1

Add

5

$5$ to both sides of the equation.

2 d = 5

$2 d = 5$

Step 2.6.2.2

Divide each term in

2 d = 5

$2 d = 5$ by

2

$2$ and simplify.

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Step 2.6.2.2.1

Divide each term in

2 d = 5

$2 d = 5$ by

2

$2$ .

\frac{2 d}{2} = \frac{5}{2}

$\frac{2 d}{2} = \frac{5}{2}$

Step 2.6.2.2.2

Simplify the left side.

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Step 2.6.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 2.6.2.2.2.1.1

Cancel the common factor.

$\frac{2 d}{2} = \frac{5}{2}$

Step 2.6.2.2.2.1.2

Divide

$d$ by

$1$ .

$d = \frac{5}{2}$

Step 2.7

The final solution is all the values that make

$(d + 1) (2 d - 5) = 0$ true.

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

Step 2.8

Next, use the negative value of the

$\pm$ to find the second solution.

$2 d^{2} - 3 d = - 5$

Step 2.9

Add

$5$ to both sides of the equation.

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

Step 2.10

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.11

Substitute the values

$a = 2$ ,

$b = - 3$ , and

$c = 5$ into the quadratic formula and solve for

$d$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Step 2.12

Simplify.

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Step 2.12.1

Simplify the numerator.

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Step 2.12.1.1

Raise

$- 3$ to the power of

$2$ .

$d = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 5}}{2 \cdot 2}$

Step 2.12.1.2

Multiply

$- 4 \cdot 2 \cdot 5$ .

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Step 2.12.1.2.1

Multiply

$- 4$ by

$2$ .

$d = \frac{3 \pm \sqrt{9 - 8 \cdot 5}}{2 \cdot 2}$

Step 2.12.1.2.2

Multiply

$- 8$ by

$5$ .

$d = \frac{3 \pm \sqrt{9 - 40}}{2 \cdot 2}$

Step 2.12.1.3

Subtract

$40$ from

$9$ .

$d = \frac{3 \pm \sqrt{- 31}}{2 \cdot 2}$

Step 2.12.1.4

Rewrite

$- 31$ as

$- 1 (31)$ .

$d = \frac{3 \pm \sqrt{- 1 \cdot 31}}{2 \cdot 2}$

Step 2.12.1.5

Rewrite

$\sqrt{- 1 (31)}$ as

$\sqrt{- 1} \cdot \sqrt{31}$ .

$d = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{31}}{2 \cdot 2}$

Step 2.12.1.6

Rewrite

$\sqrt{- 1}$ as

$i$ .

$d = \frac{3 \pm i \sqrt{31}}{2 \cdot 2}$

Step 2.12.2

Multiply

$2$ by

$2$ .

$d = \frac{3 \pm i \sqrt{31}}{4}$

Step 2.13

The final answer is the combination of both solutions.

$d = \frac{3 + i \sqrt{31}}{4}, \frac{3 - i \sqrt{31}}{4}$

Step 2.14

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

$d = - 1, \frac{5}{2}, \frac{3 + i \sqrt{31}}{4}, \frac{3 - i \sqrt{31}}{4}$