Solve for x 4x^2-4a^2x+(a^4-b^4)=0

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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

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

Step 3

Simplify.

Simplify the numerator.

Add parentheses.

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

Let $u = 4 \cdot (a^{4} - b^{4})$ . Substitute $u$ for all occurrences of $4 \cdot (a^{4} - b^{4})$ .

Apply the product rule to $- 4 a^{2}$ .

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

Raise $- 4$ to the power of $2$ .

$x = \frac{4 a^{2} \pm \sqrt{16 {(a^{2})}^{2} - 4 \cdot u}}{2 \cdot 4}$

Multiply the exponents in ${(a^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{4 a^{2} \pm \sqrt{16 a^{2 \cdot 2} - 4 \cdot u}}{2 \cdot 4}$

Multiply $2$ by $2$ .

$x = \frac{4 a^{2} \pm \sqrt{16 a^{4} - 4 \cdot u}}{2 \cdot 4}$

$x = \frac{4 a^{2} \pm \sqrt{16 a^{4} - 4 u}}{2 \cdot 4}$

Factor $4$ out of $16 a^{4} - 4 u$ .

Factor $4$ out of $16 a^{4}$ .

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

Factor $4$ out of $- 4 u$ .

$x = \frac{4 a^{2} \pm \sqrt{4 (4 a^{4}) + 4 (- u)}}{2 \cdot 4}$

Factor $4$ out of $4 (4 a^{4}) + 4 (- u)$ .

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

Replace all occurrences of $u$ with $4 \cdot (a^{4} - b^{4})$ .

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

Simplify.

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $4$ .

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

Apply the distributive property.

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

Multiply $4$ by $- 1$ .

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

Multiply $- 4$ by $- 1$ .

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

Subtract $4 a^{4}$ from $4 a^{4}$ .

$x = \frac{4 a^{2} \pm \sqrt{4 (0 + 4 b^{4})}}{2 \cdot 4}$

Add $0$ and $4 b^{4}$ .

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

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

Multiply $4$ by $4$ .

$x = \frac{4 a^{2} \pm \sqrt{16 b^{4}}}{2 \cdot 4}$

Rewrite $16 b^{4}$ as ${(4 b^{2})}^{2}$ .

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

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

$x = \frac{4 a^{2} \pm 4 b^{2}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{4 a^{2} \pm 4 b^{2}}{8}$

Simplify $\frac{4 a^{2} \pm 4 b^{2}}{8}$ .

$x = \frac{a^{2} \pm b^{2}}{2}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{a^{2} + b^{2}}{2}$

$x = \frac{(a + b) (a - b)}{2}$

Solve for x 4x^2-4a^2x+(a^4-b^4)=0

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