Basic Math Examples

$b^{2} + 6 b + a = {(b + a)}^{2}$

Step 1

Simplify ${(b + a)}^{2}$ .

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Step 1.1

Rewrite.

$b^{2} + 6 b + a = 0 + 0 + {(b + a)}^{2}$

Step 1.2

Rewrite ${(b + a)}^{2}$ as $(b + a) (b + a)$ .

$b^{2} + 6 b + a = (b + a) (b + a)$

Step 1.3

Expand $(b + a) (b + a)$ using the FOIL Method.

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Step 1.3.1

Apply the distributive property.

$b^{2} + 6 b + a = b (b + a) + a (b + a)$

Step 1.3.2

Apply the distributive property.

$b^{2} + 6 b + a = b \cdot b + b a + a (b + a)$

Step 1.3.3

Apply the distributive property.

$b^{2} + 6 b + a = b \cdot b + b a + a b + a \cdot a$

Step 1.4

Simplify and combine like terms.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Multiply $b$ by $b$ .

$b^{2} + 6 b + a = b^{2} + b a + a b + a \cdot a$

Step 1.4.1.2

Multiply $a$ by $a$ .

$b^{2} + 6 b + a = b^{2} + b a + a b + a^{2}$

Step 1.4.2

Add $b a$ and $a b$ .

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Step 1.4.2.1

Reorder $b$ and $a$ .

$b^{2} + 6 b + a = b^{2} + a b + a b + a^{2}$

Step 1.4.2.2

Add $a b$ and $a b$ .

$b^{2} + 6 b + a = b^{2} + 2 a b + a^{2}$

Step 2

Since $a$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$b^{2} + 2 a b + a^{2} = b^{2} + 6 b + a$

Step 3

Subtract $a$ from both sides of the equation.

$b^{2} + 2 a b + a^{2} - a = b^{2} + 6 b$

Step 4

Move all terms to the left side of the equation and simplify.

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Step 4.1

Move all the expressions to the left side of the equation.

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Step 4.1.1

Subtract $b^{2}$ from both sides of the equation.

$b^{2} + 2 a b + a^{2} - a - b^{2} = 6 b$

Step 4.1.2

Subtract $6 b$ from both sides of the equation.

$b^{2} + 2 a b + a^{2} - a - b^{2} - 6 b = 0$

Step 4.2

Combine the opposite terms in $b^{2} + 2 a b + a^{2} - a - b^{2} - 6 b$ .

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Step 4.2.1

Subtract $b^{2}$ from $b^{2}$ .

$2 a b + a^{2} - a + 0 - 6 b = 0$

Step 4.2.2

Add $2 a b + a^{2} - a$ and $0$ .

$2 a b + a^{2} - a - 6 b = 0$

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

Substitute the values $a = 1$ , $b = 2 b - 1$ , and $c = - 6 b$ into the quadratic formula and solve for $a$ .

$\frac{- (2 b - 1) \pm \sqrt{{(2 b - 1)}^{2} - 4 \cdot (1 \cdot (- 6 b))}}{2 \cdot 1}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Apply the distributive property.

$a = \frac{- (2 b) + 1 \pm \sqrt{{(2 b - 1)}^{2} - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.2

Multiply $2$ by $- 1$ .

$a = \frac{- 2 b + 1 \pm \sqrt{{(2 b - 1)}^{2} - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.3

Multiply $- 1$ by $- 1$ .

$a = \frac{- 2 b + 1 \pm \sqrt{{(2 b - 1)}^{2} - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.4

Rewrite ${(2 b - 1)}^{2}$ as $(2 b - 1) (2 b - 1)$ .

$a = \frac{- 2 b + 1 \pm \sqrt{(2 b - 1) (2 b - 1) - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.5

Expand $(2 b - 1) (2 b - 1)$ using the FOIL Method.

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Step 7.1.5.1

Apply the distributive property.

$a = \frac{- 2 b + 1 \pm \sqrt{2 b (2 b - 1) - 1 (2 b - 1) - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.5.2

Apply the distributive property.

$a = \frac{- 2 b + 1 \pm \sqrt{2 b (2 b) + 2 b \cdot - 1 - 1 (2 b - 1) - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.5.3

Apply the distributive property.

$a = \frac{- 2 b + 1 \pm \sqrt{2 b (2 b) + 2 b \cdot - 1 - 1 (2 b) - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6

Simplify and combine like terms.

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Step 7.1.6.1

Simplify each term.

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Step 7.1.6.1.1

Rewrite using the commutative property of multiplication.

$a = \frac{- 2 b + 1 \pm \sqrt{2 \cdot (2 b \cdot b) + 2 b \cdot - 1 - 1 (2 b) - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.1.2

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

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Step 7.1.6.1.2.1

Move $b$ .

$a = \frac{- 2 b + 1 \pm \sqrt{2 \cdot (2 (b \cdot b)) + 2 b \cdot - 1 - 1 (2 b) - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.1.2.2

Multiply $b$ by $b$ .

$a = \frac{- 2 b + 1 \pm \sqrt{2 \cdot (2 b^{2}) + 2 b \cdot - 1 - 1 (2 b) - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.1.3

Multiply $2$ by $2$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} + 2 b \cdot - 1 - 1 (2 b) - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.1.4

Multiply $- 1$ by $2$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} - 2 b - 1 (2 b) - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.1.5

Multiply $2$ by $- 1$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} - 2 b - 2 b - 1 \cdot - 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.1.6

Multiply $- 1$ by $- 1$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} - 2 b - 2 b + 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.6.2

Subtract $2 b$ from $- 2 b$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} - 4 b + 1 - 4 \cdot 1 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.7

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 7.1.7.1

Multiply $- 4$ by $1$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} - 4 b + 1 - 4 \cdot (- 6 b)}}{2 \cdot 1}$

Step 7.1.7.2

Multiply $- 4$ by $- 6$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} - 4 b + 1 + 24 b}}{2 \cdot 1}$

Step 7.1.8

Add $- 4 b$ and $24 b$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} + 20 b + 1}}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$a = \frac{- 2 b + 1 \pm \sqrt{4 b^{2} + 20 b + 1}}{2}$

Step 8

The final answer is the combination of both solutions.

$a = - \frac{2 b - 1 - \sqrt{4 b^{2} + 20 b + 1}}{2}$

$a = - \frac{2 b - 1 + \sqrt{4 b^{2} + 20 b + 1}}{2}$