Solve for r square root of r+1=2r

$\sqrt{r} + 1 = 2 r$

Step 1

Subtract $1$ from both sides of the equation.

$\sqrt{r} = 2 r - 1$

Step 2

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

${\sqrt{r}}^{2} = {(2 r - 1)}^{2}$

Step 3

Simplify each side of the equation.

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

${(r^{\frac{1}{2}})}^{2} = {(2 r - 1)}^{2}$

Simplify the left side.

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

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

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

$r^{\frac{1}{2} \cdot 2} = {(2 r - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$r^{\frac{1}{2} \cdot 2} = {(2 r - 1)}^{2}$

Rewrite the expression.

$r^{1} = {(2 r - 1)}^{2}$

Simplify.

$r = {(2 r - 1)}^{2}$

Simplify the right side.

Simplify ${(2 r - 1)}^{2}$ .

Rewrite ${(2 r - 1)}^{2}$ as $(2 r - 1) (2 r - 1)$ .

$r = (2 r - 1) (2 r - 1)$

Expand $(2 r - 1) (2 r - 1)$ using the FOIL Method.

Apply the distributive property.

$r = 2 r (2 r - 1) - 1 (2 r - 1)$

Apply the distributive property.

$r = 2 r (2 r) + 2 r \cdot - 1 - 1 (2 r - 1)$

Apply the distributive property.

$r = 2 r (2 r) + 2 r \cdot - 1 - 1 (2 r) - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$r = 2 \cdot 2 r \cdot r + 2 r \cdot - 1 - 1 (2 r) - 1 \cdot - 1$

Multiply $r$ by $r$ by adding the exponents.

Move $r$ .

$r = 2 \cdot 2 (r \cdot r) + 2 r \cdot - 1 - 1 (2 r) - 1 \cdot - 1$

Multiply $r$ by $r$ .

$r = 2 \cdot 2 r^{2} + 2 r \cdot - 1 - 1 (2 r) - 1 \cdot - 1$

Multiply $2$ by $2$ .

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

Multiply $- 1$ by $2$ .

$r = 4 r^{2} - 2 r - 1 (2 r) - 1 \cdot - 1$

Multiply $2$ by $- 1$ .

$r = 4 r^{2} - 2 r - 2 r - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

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

Subtract $2 r$ from $- 2 r$ .

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

Step 4

Solve for $r$ .

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

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

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

Subtract $r$ from both sides of the equation.

$4 r^{2} - 4 r + 1 - r = 0$

Subtract $r$ from $- 4 r$ .

$4 r^{2} - 5 r + 1 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 1 = 4$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 r$ .

$4 r^{2} - 5 r + 1 = 0$

Rewrite $- 5$ as $- 1$ plus $- 4$

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

Apply the distributive property.

$4 r^{2} - 1 r - 4 r + 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

$r (4 r - 1) - (4 r - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 r - 1$ .

$(4 r - 1) (r - 1) = 0$

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

$4 r - 1 = 0$

$r - 1 = 0$

Set $4 r - 1$ equal to $0$ and solve for $r$ .

Set $4 r - 1$ equal to $0$ .

$4 r - 1 = 0$

Solve $4 r - 1 = 0$ for $r$ .

Add $1$ to both sides of the equation.

$4 r = 1$

Divide each term in $4 r = 1$ by $4$ and simplify.

Divide each term in $4 r = 1$ by $4$ .

$\frac{4 r}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 r}{4} = \frac{1}{4}$

Divide $r$ by $1$ .

$r = \frac{1}{4}$

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

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

$r - 1 = 0$

Add $1$ to both sides of the equation.

$r = 1$

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

$r = \frac{1}{4}, 1$

Step 5

Exclude the solutions that do not make $\sqrt{r} + 1 = 2 r$ true.

$r = 1$

Solve for r square root of r+1=2r

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