Solve for t t = square root of t^2+2t+4

$t = \sqrt{t^{2} + 2 t + 4}$

Step 1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{t^{2} + 2 t + 4} = t$

Step 2

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

${\sqrt{t^{2} + 2 t + 4}}^{2} = t^{2}$

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$t^{2} + 2 t + 4 = t^{2}$

Step 4

Solve for $t$ .

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

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

$t^{2} + 2 t + 4 - t^{2} = 0$

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

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

$2 t + 4 + 0 = 0$

Add $2 t + 4$ and $0$ .

$2 t + 4 = 0$

Subtract $4$ from both sides of the equation.

$2 t = - 4$

Divide each term in $2 t = - 4$ by $2$ and simplify.

Divide each term in $2 t = - 4$ by $2$ .

$\frac{2 t}{2} = \frac{- 4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 t}{2} = \frac{- 4}{2}$

Divide $t$ by $1$ .

$t = \frac{- 4}{2}$

Simplify the right side.

Divide $- 4$ by $2$ .

$t = - 2$

Step 5

Exclude the solutions that do not make $t = \sqrt{t^{2} + 2 t + 4}$ true.

$No solution$

Solve for t t = square root of t^2+2t+4

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