Solve Using the Square Root Property y^2-4y+4=-64

$y^{2} - 4 y + 4 = - 64$

Step 1

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

Add $64$ to both sides of the equation.

$y^{2} - 4 y + 4 + 64 = 0$

Add $4$ and $64$ .

$y^{2} - 4 y + 68 = 0$

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 $a = 1$ , $b = - 4$ , and $c = 68$ into the quadratic formula and solve for $y$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 68)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 68}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 68$ .

Multiply $- 4$ by $1$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 68}}{2 \cdot 1}$

Multiply $- 4$ by $68$ .

$y = \frac{4 \pm \sqrt{16 - 272}}{2 \cdot 1}$

Subtract $272$ from $16$ .

$y = \frac{4 \pm \sqrt{- 256}}{2 \cdot 1}$

Rewrite $- 256$ as $- 1 (256)$ .

$y = \frac{4 \pm \sqrt{- 1 \cdot 256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (256)}$ as $\sqrt{- 1} \cdot \sqrt{256}$ .

$y = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{4 \pm i \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$y = \frac{4 \pm i \cdot \sqrt{16^{2}}}{2 \cdot 1}$

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

$y = \frac{4 \pm i \cdot 16}{2 \cdot 1}$

Move $16$ to the left of $i$ .

$y = \frac{4 \pm 16 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{4 \pm 16 i}{2}$

Simplify $\frac{4 \pm 16 i}{2}$ .

$y = 2 \pm 8 i$

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 68}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 68$ .

Multiply $- 4$ by $1$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 68}}{2 \cdot 1}$

Multiply $- 4$ by $68$ .

$y = \frac{4 \pm \sqrt{16 - 272}}{2 \cdot 1}$

Subtract $272$ from $16$ .

$y = \frac{4 \pm \sqrt{- 256}}{2 \cdot 1}$

Rewrite $- 256$ as $- 1 (256)$ .

$y = \frac{4 \pm \sqrt{- 1 \cdot 256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (256)}$ as $\sqrt{- 1} \cdot \sqrt{256}$ .

$y = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{4 \pm i \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$y = \frac{4 \pm i \cdot \sqrt{16^{2}}}{2 \cdot 1}$

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

$y = \frac{4 \pm i \cdot 16}{2 \cdot 1}$

Move $16$ to the left of $i$ .

$y = \frac{4 \pm 16 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{4 \pm 16 i}{2}$

Simplify $\frac{4 \pm 16 i}{2}$ .

$y = 2 \pm 8 i$

Change the $\pm$ to $+$ .

$y = 2 + 8 i$

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 68}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 68$ .

Multiply $- 4$ by $1$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 68}}{2 \cdot 1}$

Multiply $- 4$ by $68$ .

$y = \frac{4 \pm \sqrt{16 - 272}}{2 \cdot 1}$

Subtract $272$ from $16$ .

$y = \frac{4 \pm \sqrt{- 256}}{2 \cdot 1}$

Rewrite $- 256$ as $- 1 (256)$ .

$y = \frac{4 \pm \sqrt{- 1 \cdot 256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (256)}$ as $\sqrt{- 1} \cdot \sqrt{256}$ .

$y = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{4 \pm i \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$y = \frac{4 \pm i \cdot \sqrt{16^{2}}}{2 \cdot 1}$

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

$y = \frac{4 \pm i \cdot 16}{2 \cdot 1}$

Move $16$ to the left of $i$ .

$y = \frac{4 \pm 16 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{4 \pm 16 i}{2}$

Simplify $\frac{4 \pm 16 i}{2}$ .

$y = 2 \pm 8 i$

Change the $\pm$ to $-$ .

$y = 2 - 8 i$

Step 7

The final answer is the combination of both solutions.

$y = 2 + 8 i, 2 - 8 i$

Solve Using the Square Root Property y^2-4y+4=-64

Related Questions

Support LCMGCF.com by disabling your adblocker.