Algebra Examples

$4 x^{2} + 5 b x = - b^{2}$

Step 1

Add $b^{2}$ to both sides of the equation.

$4 x^{2} + 5 b x + b^{2} = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 4$ , $b = 5 b$ , and $c = b^{2}$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Rewrite $4 \cdot 4 \cdot b^{2}$ as ${(2 \cdot 2 b)}^{2}$ .

$x = \frac{- 5 b \pm \sqrt{{(5 b)}^{2} - {(2 \cdot (2 b))}^{2}}}{2 \cdot 4}$

Step 4.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5 b$ and $b = 2 \cdot 2 b$ .

$x = \frac{- 5 b \pm \sqrt{(5 b + 2 \cdot (2 b)) (5 b - (2 \cdot (2 b)))}}{2 \cdot 4}$

Step 4.1.3

Simplify.

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Step 4.1.3.1

Factor $b$ out of $5 b + 2 \cdot 2 b$ .

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Step 4.1.3.1.1

Factor $b$ out of $5 b$ .

$x = \frac{- 5 b \pm \sqrt{(b \cdot 5 + 2 \cdot (2 b)) (5 b - (2 \cdot (2 b)))}}{2 \cdot 4}$

Step 4.1.3.1.2

Factor $b$ out of $2 \cdot 2 b$ .

$x = \frac{- 5 b \pm \sqrt{(b \cdot 5 + b (2 \cdot 2)) (5 b - (2 \cdot (2 b)))}}{2 \cdot 4}$

Step 4.1.3.1.3

Factor $b$ out of $b \cdot 5 + b (2 \cdot 2)$ .

$x = \frac{- 5 b \pm \sqrt{b (5 + 2 \cdot 2) (5 b - (2 \cdot (2 b)))}}{2 \cdot 4}$

Step 4.1.3.2

Multiply $2$ by $2$ .

$x = \frac{- 5 b \pm \sqrt{b (5 + 4) (5 b - (2 \cdot (2 b)))}}{2 \cdot 4}$

Step 4.1.3.3

Add $5$ and $4$ .

$x = \frac{- 5 b \pm \sqrt{b \cdot (9 (5 b - (2 \cdot (2 b))))}}{2 \cdot 4}$

Step 4.1.3.4

Combine exponents.

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Step 4.1.3.4.1

Multiply $2$ by $2$ .

$x = \frac{- 5 b \pm \sqrt{b \cdot (9 (5 b - (4 b)))}}{2 \cdot 4}$

Step 4.1.3.4.2

Multiply $4$ by $- 1$ .

$x = \frac{- 5 b \pm \sqrt{b \cdot (9 (5 b - 4 b))}}{2 \cdot 4}$

Step 4.1.4

Subtract $4 b$ from $5 b$ .

$x = \frac{- 5 b \pm \sqrt{b \cdot (9 b)}}{2 \cdot 4}$

Step 4.1.5

Combine exponents.

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Step 4.1.5.1

Raise $b$ to the power of $1$ .

$x = \frac{- 5 b \pm \sqrt{9 (b \cdot b)}}{2 \cdot 4}$

Step 4.1.5.2

Raise $b$ to the power of $1$ .

$x = \frac{- 5 b \pm \sqrt{9 (b \cdot b)}}{2 \cdot 4}$

Step 4.1.5.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \frac{- 5 b \pm \sqrt{9 b^{1 + 1}}}{2 \cdot 4}$

Step 4.1.5.4

Add $1$ and $1$ .

$x = \frac{- 5 b \pm \sqrt{9 b^{2}}}{2 \cdot 4}$

Step 4.1.6

Rewrite $9 b^{2}$ as ${(3 b)}^{2}$ .

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

Step 4.1.7

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

$x = \frac{- 5 b \pm 3 b}{2 \cdot 4}$

Step 4.2

Multiply $2$ by $4$ .

$x = \frac{- 5 b \pm 3 b}{8}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{b}{4}$

$x = - b$