Solve for q 2q = square root of 10q-6

$2 q = \sqrt{10 q - 6}$

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

$\sqrt{10 q - 6} = 2 q$

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

${\sqrt{10 q - 6}}^{2} = {(2 q)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10 q - 6}$ as ${(10 q - 6)}^{\frac{1}{2}}$ .

${({(10 q - 6)}^{\frac{1}{2}})}^{2} = {(2 q)}^{2}$

Simplify the left side.

Simplify ${({(10 q - 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(10 q - 6)}^{\frac{1}{2}})}^{2}$ .

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

${(10 q - 6)}^{\frac{1}{2} \cdot 2} = {(2 q)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(10 q - 6)}^{\frac{1}{2} \cdot 2} = {(2 q)}^{2}$

Rewrite the expression.

${(10 q - 6)}^{1} = {(2 q)}^{2}$

Simplify.

$10 q - 6 = {(2 q)}^{2}$

Simplify the right side.

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

Apply the product rule to $2 q$ .

$10 q - 6 = 2^{2} q^{2}$

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

$10 q - 6 = 4 q^{2}$

Solve for $q$ .

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

$10 q - 6 - 4 q^{2} = 0$

Factor the left side of the equation.

Factor $- 2$ out of $10 q - 6 - 4 q^{2}$ .

Reorder the expression.

Move $- 6$ .

$10 q - 4 q^{2} - 6 = 0$

Reorder $10 q$ and $- 4 q^{2}$ .

$- 4 q^{2} + 10 q - 6 = 0$

Factor $- 2$ out of $- 4 q^{2}$ .

$- 2 (2 q^{2}) + 10 q - 6 = 0$

Factor $- 2$ out of $10 q$ .

$- 2 (2 q^{2}) - 2 (- 5 q) - 6 = 0$

Factor $- 2$ out of $- 6$ .

$- 2 (2 q^{2}) - 2 (- 5 q) - 2 \cdot 3 = 0$

Factor $- 2$ out of $- 2 (2 q^{2}) - 2 (- 5 q)$ .

$- 2 (2 q^{2} - 5 q) - 2 \cdot 3 = 0$

Factor $- 2$ out of $- 2 (2 q^{2} - 5 q) - 2 (3)$ .

$- 2 (2 q^{2} - 5 q + 3) = 0$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 q$ .

$- 2 (2 q^{2} - 5 q + 3) = 0$

Rewrite $- 5$ as $- 2$ plus $- 3$

$- 2 (2 q^{2} + (- 2 - 3) q + 3) = 0$

Apply the distributive property.

$- 2 (2 q^{2} - 2 q - 3 q + 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 2 ((2 q^{2} - 2 q) - 3 q + 3) = 0$

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

$- 2 (2 q (q - 1) - 3 (q - 1)) = 0$

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

$- 2 ((q - 1) (2 q - 3)) = 0$

Remove unnecessary parentheses.

$- 2 (q - 1) (2 q - 3) = 0$

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

$q - 1 = 0$

$2 q - 3 = 0$

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

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

$q - 1 = 0$

Add $1$ to both sides of the equation.

$q = 1$

Set $2 q - 3$ equal to $0$ and solve for $q$ .

Set $2 q - 3$ equal to $0$ .

$2 q - 3 = 0$

Solve $2 q - 3 = 0$ for $q$ .

Add $3$ to both sides of the equation.

$2 q = 3$

Divide each term in $2 q = 3$ by $2$ and simplify.

Divide each term in $2 q = 3$ by $2$ .

$\frac{2 q}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 q}{2} = \frac{3}{2}$

Divide $q$ by $1$ .

$q = \frac{3}{2}$

The final solution is all the values that make $- 2 (q - 1) (2 q - 3) = 0$ true.

$q = 1, \frac{3}{2}$

The result can be shown in multiple forms.

Exact Form:

$q = 1, \frac{3}{2}$

Decimal Form:

$q = 1, 1.5$

Mixed Number Form:

$q = 1, 1 \frac{1}{2}$

Solve for q 2q = square root of 10q-6

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