Basic Math Examples

$a (a - c) + 4 b (c - 4 b) = 0$

Step 1

Simplify each term.

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Step 1.1

Apply the distributive property.

$a \cdot a + a (- c) + 4 b (c - 4 b) = 0$

Step 1.2

Multiply $a$ by $a$ .

$a^{2} + a (- c) + 4 b (c - 4 b) = 0$

Step 1.3

Rewrite using the commutative property of multiplication.

$a^{2} - a c + 4 b (c - 4 b) = 0$

Step 1.4

Apply the distributive property.

$a^{2} - a c + 4 b c + 4 b (- 4 b) = 0$

Step 1.5

Rewrite using the commutative property of multiplication.

$a^{2} - a c + 4 b c + 4 \cdot - 4 b \cdot b = 0$

Step 1.6

Simplify each term.

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Step 1.6.1

Multiply $b$ by $b$ by adding the exponents.

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Step 1.6.1.1

Move $b$ .

$a^{2} - a c + 4 b c + 4 \cdot - 4 (b \cdot b) = 0$

Step 1.6.1.2

Multiply $b$ by $b$ .

$a^{2} - a c + 4 b c + 4 \cdot - 4 b^{2} = 0$

Step 1.6.2

Multiply $4$ by $- 4$ .

$a^{2} - a c + 4 b c - 16 b^{2} = 0$

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 $a = 1$ , $b = - c$ , and $c = 4 b c - 16 b^{2}$ into the quadratic formula and solve for $a$ .

$\frac{c \pm \sqrt{{(- c)}^{2} - 4 \cdot (1 \cdot (4 b c - 16 b^{2}))}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the product rule to $- c$ .

$a = \frac{c \pm \sqrt{{(- 1)}^{2} c^{2} - 4 \cdot 1 \cdot (4 b c - 16 b^{2})}}{2 \cdot 1}$

Step 4.1.2

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

$a = \frac{c \pm \sqrt{1 c^{2} - 4 \cdot 1 \cdot (4 b c - 16 b^{2})}}{2 \cdot 1}$

Step 4.1.3

Multiply $c^{2}$ by $1$ .

$a = \frac{c \pm \sqrt{c^{2} - 4 \cdot 1 \cdot (4 b c - 16 b^{2})}}{2 \cdot 1}$

Step 4.1.4

Multiply $- 4$ by $1$ .

$a = \frac{c \pm \sqrt{c^{2} - 4 \cdot (4 b c - 16 b^{2})}}{2 \cdot 1}$

Step 4.1.5

Apply the distributive property.

$a = \frac{c \pm \sqrt{c^{2} - 4 (4 b c) - 4 (- 16 b^{2})}}{2 \cdot 1}$

Step 4.1.6

Multiply $4$ by $- 4$ .

$a = \frac{c \pm \sqrt{c^{2} - 16 (b c) - 4 (- 16 b^{2})}}{2 \cdot 1}$

Step 4.1.7

Multiply $- 16$ by $- 4$ .

$a = \frac{c \pm \sqrt{c^{2} - 16 (b c) + 64 b^{2}}}{2 \cdot 1}$

Step 4.1.8

Factor using the perfect square rule.

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Step 4.1.8.1

Rewrite $64 b^{2}$ as ${(8 b)}^{2}$ .

$a = \frac{c \pm \sqrt{c^{2} - 16 b c + {(8 b)}^{2}}}{2 \cdot 1}$

Step 4.1.8.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 b c = 2 \cdot c \cdot (8 b)$

Step 4.1.8.3

Rewrite the polynomial.

$a = \frac{c \pm \sqrt{c^{2} - 2 \cdot c \cdot (8 b) + {(8 b)}^{2}}}{2 \cdot 1}$

Step 4.1.8.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = c$ and $b = 8 b$ .

$a = \frac{c \pm \sqrt{{(c - 8 b)}^{2}}}{2 \cdot 1}$

Step 4.1.9

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

$a = \frac{c \pm (c - 8 b)}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$a = \frac{c \pm (c - 8 b)}{2}$

Step 5

The final answer is the combination of both solutions.

$a = c - 4 b$

$a = 4 b$