Solve for I s=2 square root of I-9

$s = 2 \sqrt{I} - 9$

Step 1

Rewrite the equation as $2 \sqrt{I} - 9 = s$ .

$2 \sqrt{I} - 9 = s$

Step 2

Add $9$ to both sides of the equation.

$2 \sqrt{I} = s + 9$

Step 3

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

${(2 \sqrt{I})}^{2} = {(s + 9)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{I}$ as $I^{\frac{1}{2}}$ .

${(2 I^{\frac{1}{2}})}^{2} = {(s + 9)}^{2}$

Simplify the left side.

Simplify ${(2 I^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 I^{\frac{1}{2}}$ .

$2^{2} {(I^{\frac{1}{2}})}^{2} = {(s + 9)}^{2}$

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

$4 {(I^{\frac{1}{2}})}^{2} = {(s + 9)}^{2}$

Multiply the exponents in ${(I^{\frac{1}{2}})}^{2}$ .

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

$4 I^{\frac{1}{2} \cdot 2} = {(s + 9)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 I^{\frac{1}{2} \cdot 2} = {(s + 9)}^{2}$

Rewrite the expression.

$4 I^{1} = {(s + 9)}^{2}$

Simplify.

$4 I = {(s + 9)}^{2}$

Simplify the right side.

Simplify ${(s + 9)}^{2}$ .

Rewrite ${(s + 9)}^{2}$ as $(s + 9) (s + 9)$ .

$4 I = (s + 9) (s + 9)$

Expand $(s + 9) (s + 9)$ using the FOIL Method.

Apply the distributive property.

$4 I = s (s + 9) + 9 (s + 9)$

Apply the distributive property.

$4 I = s \cdot s + s \cdot 9 + 9 (s + 9)$

Apply the distributive property.

$4 I = s \cdot s + s \cdot 9 + 9 s + 9 \cdot 9$

Simplify and combine like terms.

Simplify each term.

Multiply $s$ by $s$ .

$4 I = s^{2} + s \cdot 9 + 9 s + 9 \cdot 9$

Move $9$ to the left of $s$ .

$4 I = s^{2} + 9 \cdot s + 9 s + 9 \cdot 9$

Multiply $9$ by $9$ .

$4 I = s^{2} + 9 s + 9 s + 81$

Add $9 s$ and $9 s$ .

$4 I = s^{2} + 18 s + 81$

Step 5

Divide each term in $4 I = s^{2} + 18 s + 81$ by $4$ and simplify.

Divide each term in $4 I = s^{2} + 18 s + 81$ by $4$ .

$\frac{4 I}{4} = \frac{s^{2}}{4} + \frac{18 s}{4} + \frac{81}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 I}{4} = \frac{s^{2}}{4} + \frac{18 s}{4} + \frac{81}{4}$

Divide $I$ by $1$ .

$I = \frac{s^{2}}{4} + \frac{18 s}{4} + \frac{81}{4}$

Simplify the right side.

Cancel the common factor of $18$ and $4$ .

Factor $2$ out of $18 s$ .

$I = \frac{s^{2}}{4} + \frac{2 (9 s)}{4} + \frac{81}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$I = \frac{s^{2}}{4} + \frac{2 (9 s)}{2 (2)} + \frac{81}{4}$

Cancel the common factor.

$I = \frac{s^{2}}{4} + \frac{2 (9 s)}{2 \cdot 2} + \frac{81}{4}$

Rewrite the expression.

$I = \frac{s^{2}}{4} + \frac{9 s}{2} + \frac{81}{4}$

Solve for I s=2 square root of I-9

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