Solve for y -4=2- square root of y^2-4y+15

$- 4 = 2 - \sqrt{y^{2} - 4 y + 15}$

Step 1

Rewrite the equation as $2 - \sqrt{y^{2} - 4 y + 15} = - 4$ .

$2 - \sqrt{y^{2} - 4 y + 15} = - 4$

Step 2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y^{2} - 4 y + 15}$ as ${(y^{2} - 4 y + 15)}^{\frac{1}{2}}$ .

$2 - {(y^{2} - 4 y + 15)}^{\frac{1}{2}} = - 4$

Step 3

Move all terms not containing $y$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$- {(y^{2} - 4 y + 15)}^{\frac{1}{2}} = - 4 - 2$

Subtract $2$ from $- 4$ .

$- {(y^{2} - 4 y + 15)}^{\frac{1}{2}} = - 6$

Step 4

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

${(- {(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2} = {(- 6)}^{2}$

Step 5

Simplify the exponent.

Simplify the left side.

Simplify ${(- {(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(y^{2} - 4 y + 15)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2} = {(- 6)}^{2}$

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

$1 {({(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2} = {(- 6)}^{2}$

Multiply ${({(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2} = {(- 6)}^{2}$

Multiply the exponents in ${({(y^{2} - 4 y + 15)}^{\frac{1}{2}})}^{2}$ .

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

${(y^{2} - 4 y + 15)}^{\frac{1}{2} \cdot 2} = {(- 6)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y^{2} - 4 y + 15)}^{\frac{1}{2} \cdot 2} = {(- 6)}^{2}$

Rewrite the expression.

${(y^{2} - 4 y + 15)}^{1} = {(- 6)}^{2}$

Simplify.

$y^{2} - 4 y + 15 = {(- 6)}^{2}$

Simplify the right side.

Raise $- 6$ to the power of $2$ .

$y^{2} - 4 y + 15 = 36$

Step 6

Solve for $y$ .

Subtract $36$ from both sides of the equation.

$y^{2} - 4 y + 15 - 36 = 0$

Subtract $36$ from $15$ .

$y^{2} - 4 y - 21 = 0$

Factor $y^{2} - 4 y - 21$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 21$ and whose sum is $- 4$ .

$- 7, 3$

Write the factored form using these integers.

$(y - 7) (y + 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 7 = 0$

$y + 3 = 0$

Set $y - 7$ equal to $0$ and solve for $y$ .

Set $y - 7$ equal to $0$ .

$y - 7 = 0$

Add $7$ to both sides of the equation.

$y = 7$

Set $y + 3$ equal to $0$ and solve for $y$ .

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

Subtract $3$ from both sides of the equation.

$y = - 3$

The final solution is all the values that make $(y - 7) (y + 3) = 0$ true.

$y = 7, - 3$

Solve for y -4=2- square root of y^2-4y+15

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