Solve for y y^4-17y^2=-16

$y^{4} - 17 y^{2} = - 16$

Step 1

Add $16$ to both sides of the equation.

$y^{4} - 17 y^{2} + 16 = 0$

Step 2

Factor the left side of the equation.

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

${(y^{2})}^{2} - 17 y^{2} + 16 = 0$

Let $u = y^{2}$ . Substitute $u$ for all occurrences of $y^{2}$ .

$u^{2} - 17 u + 16 = 0$

Factor $u^{2} - 17 u + 16$ 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 $16$ and whose sum is $- 17$ .

$- 16, - 1$

Write the factored form using these integers.

$(u - 16) (u - 1) = 0$

Replace all occurrences of $u$ with $y^{2}$ .

$(y^{2} - 16) (y^{2} - 1) = 0$

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

$(y^{2} - 4^{2}) (y^{2} - 1) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 4$ .

$(y + 4) (y - 4) (y^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(y + 4) (y - 4) (y^{2} - 1^{2}) = 0$

Factor.

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

$(y + 4) (y - 4) ((y + 1) (y - 1)) = 0$

Remove unnecessary parentheses.

$(y + 4) (y - 4) (y + 1) (y - 1) = 0$

Step 3

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

$y + 4 = 0$

$y - 4 = 0$

$y + 1 = 0$

$y - 1 = 0$

Step 4

Set $y + 4$ equal to $0$ and solve for $y$ .

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

Subtract $4$ from both sides of the equation.

$y = - 4$

Step 5

Set $y - 4$ equal to $0$ and solve for $y$ .

Set $y - 4$ equal to $0$ .

$y - 4 = 0$

Add $4$ to both sides of the equation.

$y = 4$

Step 6

Set $y + 1$ equal to $0$ and solve for $y$ .

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 7

Set $y - 1$ equal to $0$ and solve for $y$ .

Set $y - 1$ equal to $0$ .

$y - 1 = 0$

Add $1$ to both sides of the equation.

$y = 1$

Step 8

The final solution is all the values that make $(y + 4) (y - 4) (y + 1) (y - 1) = 0$ true.

$y = - 4, 4, - 1, 1$

Solve for y y^4-17y^2=-16

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