Solve for a aa+bb=6ab

$a \cdot a + b \cdot b = 6 a b$

Step 1

Simplify each term.

Multiply $a$ by $a$ .

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

Multiply $b$ by $b$ .

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

Step 2

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

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

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

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

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

Step 5

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 1 \cdot b^{2}$ as ${(2 \cdot 1 b)}^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 6 b$ and $b = 2 \cdot 1 b$ .

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

Simplify.

Factor $2 b$ out of $- 6 b + 2 \cdot 1 b$ .

Factor $2 b$ out of $- 6 b$ .

$a = \frac{6 b \pm \sqrt{(2 b (- 3) + 2 \cdot (1 b)) (- 6 b - (2 \cdot (1 b)))}}{2 \cdot 1}$

Factor $2 b$ out of $2 \cdot 1 b$ .

$a = \frac{6 b \pm \sqrt{(2 b (- 3) + 2 b (1)) (- 6 b - (2 \cdot (1 b)))}}{2 \cdot 1}$

Factor $2 b$ out of $2 b (- 3) + 2 b (1)$ .

$a = \frac{6 b \pm \sqrt{2 b (- 3 + 1) (- 6 b - (2 \cdot (1 b)))}}{2 \cdot 1}$

Add $- 3$ and $1$ .

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

Multiply $- 2$ by $2$ .

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

Combine exponents.

Multiply $2$ by $1$ .

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

Multiply $2$ by $- 1$ .

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

Subtract $2 b$ from $- 6 b$ .

$a = \frac{6 b \pm \sqrt{- 4 b \cdot (- 8 b)}}{2 \cdot 1}$

Combine exponents.

Multiply $- 8$ by $- 4$ .

$a = \frac{6 b \pm \sqrt{32 b \cdot b}}{2 \cdot 1}$

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

$a = \frac{6 b \pm \sqrt{32 (b \cdot b)}}{2 \cdot 1}$

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

$a = \frac{6 b \pm \sqrt{32 (b \cdot b)}}{2 \cdot 1}$

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

$a = \frac{6 b \pm \sqrt{32 b^{1 + 1}}}{2 \cdot 1}$

Add $1$ and $1$ .

$a = \frac{6 b \pm \sqrt{32 b^{2}}}{2 \cdot 1}$

Rewrite $32 b^{2}$ as ${(4 b)}^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$a = \frac{6 b \pm \sqrt{16 (2) b^{2}}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Move $2$ .

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

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

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

Pull terms out from under the radical.

$a = \frac{6 b \pm 4 b \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{6 b \pm 4 b \sqrt{2}}{2}$

Simplify $\frac{6 b \pm 4 b \sqrt{2}}{2}$ .

$a = 3 b \pm 2 b \sqrt{2}$

Step 6

The final answer is the combination of both solutions.

$a = 3 b + 2 b \sqrt{2}$

$a = 3 b - 2 b \sqrt{2}$

Solve for a a*a+b*b=6ab

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Solve for a aa+bb=6ab