Solve for a square root of 1-3a=2a

$\sqrt{1 - 3 a} = 2 a$

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

${\sqrt{1 - 3 a}}^{2} = {(2 a)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 - 3 a}$ as ${(1 - 3 a)}^{\frac{1}{2}}$ .

${({(1 - 3 a)}^{\frac{1}{2}})}^{2} = {(2 a)}^{2}$

Simplify the left side.

Simplify ${({(1 - 3 a)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(1 - 3 a)}^{\frac{1}{2}})}^{2}$ .

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

${(1 - 3 a)}^{\frac{1}{2} \cdot 2} = {(2 a)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(1 - 3 a)}^{\frac{1}{2} \cdot 2} = {(2 a)}^{2}$

Rewrite the expression.

${(1 - 3 a)}^{1} = {(2 a)}^{2}$

Simplify.

$1 - 3 a = {(2 a)}^{2}$

Simplify the right side.

Simplify ${(2 a)}^{2}$ .

Apply the product rule to $2 a$ .

$1 - 3 a = 2^{2} a^{2}$

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

$1 - 3 a = 4 a^{2}$

Solve for $a$ .

Subtract $4 a^{2}$ from both sides of the equation.

$1 - 3 a - 4 a^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $1 - 3 a - 4 a^{2}$ .

Reorder the expression.

Move $1$ .

$- 3 a - 4 a^{2} + 1 = 0$

Reorder $- 3 a$ and $- 4 a^{2}$ .

$- 4 a^{2} - 3 a + 1 = 0$

Factor $- 1$ out of $- 4 a^{2}$ .

$- (4 a^{2}) - 3 a + 1 = 0$

Factor $- 1$ out of $- 3 a$ .

$- (4 a^{2}) - (3 a) + 1 = 0$

Rewrite $1$ as $- 1 (- 1)$ .

$- (4 a^{2}) - (3 a) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (4 a^{2}) - (3 a)$ .

$- (4 a^{2} + 3 a) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (4 a^{2} + 3 a) - 1 (- 1)$ .

$- (4 a^{2} + 3 a - 1) = 0$

Factor.

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 $b = 3$ .

Factor $3$ out of $3 a$ .

$- (4 a^{2} + 3 (a) - 1) = 0$

Rewrite $3$ as $- 1$ plus $4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

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

$4 a - 1 = 0$

$a + 1 = 0$

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

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

$4 a - 1 = 0$

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

Add $1$ to both sides of the equation.

$4 a = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $a$ by $1$ .

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

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

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

$a + 1 = 0$

Subtract $1$ from both sides of the equation.

$a = - 1$

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

$a = \frac{1}{4}, - 1$

Exclude the solutions that do not make $\sqrt{1 - 3 a} = 2 a$ true.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$a = 0.25$

Solve for a square root of 1-3a=2a

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