Solve for y square root of y^4+2y^2+1=y^2+1

$\sqrt{y^{4} + 2 y^{2} + 1} = y^{2} + 1$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{y^{4} + 2 y^{2} + 1}}^{2} = {(y^{2} + 1)}^{2}$

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(y^{2} + 1)}^{2}$ .

Rewrite ${(y^{2} + 1)}^{2}$ as $(y^{2} + 1) (y^{2} + 1)$ .

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

Expand $(y^{2} + 1) (y^{2} + 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$y^{4} + 2 y^{2} + 1 = y^{2} y^{2} + y^{2} \cdot 1 + 1 (y^{2} + 1)$

Apply the distributive property.

$y^{4} + 2 y^{2} + 1 = y^{2} y^{2} + y^{2} \cdot 1 + 1 y^{2} + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

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

$y^{4} + 2 y^{2} + 1 = y^{2 + 2} + y^{2} \cdot 1 + 1 y^{2} + 1 \cdot 1$

Add $2$ and $2$ .

$y^{4} + 2 y^{2} + 1 = y^{4} + y^{2} \cdot 1 + 1 y^{2} + 1 \cdot 1$

Multiply $y^{2}$ by $1$ .

$y^{4} + 2 y^{2} + 1 = y^{4} + y^{2} + 1 y^{2} + 1 \cdot 1$

Multiply $y^{2}$ by $1$ .

$y^{4} + 2 y^{2} + 1 = y^{4} + y^{2} + y^{2} + 1 \cdot 1$

Multiply $1$ by $1$ .

$y^{4} + 2 y^{2} + 1 = y^{4} + y^{2} + y^{2} + 1$

Add $y^{2}$ and $y^{2}$ .

$y^{4} + 2 y^{2} + 1 = y^{4} + 2 y^{2} + 1$

Step 3

Solve for $y$ .

Move all terms containing $y$ to the left side of the equation.

Subtract $y^{4}$ from both sides of the equation.

$y^{4} + 2 y^{2} + 1 - y^{4} = 2 y^{2} + 1$

Subtract $2 y^{2}$ from both sides of the equation.

$y^{4} + 2 y^{2} + 1 - y^{4} - 2 y^{2} = 1$

Combine the opposite terms in $y^{4} + 2 y^{2} + 1 - y^{4} - 2 y^{2}$ .

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

$2 y^{2} + 1 + 0 - 2 y^{2} = 1$

Add $2 y^{2} + 1$ and $0$ .

$2 y^{2} + 1 - 2 y^{2} = 1$

Subtract $2 y^{2}$ from $2 y^{2}$ .

$0 + 1 = 1$

Add $0$ and $1$ .

$1 = 1$

Since $1 = 1$ , the equation will always be true for any value of $y$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for y square root of y^4+2y^2+1=y^2+1

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