Solve for x 2x = square root of 17x-15

$2 x = \sqrt{17 x - 15}$

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.

$\sqrt{17 x - 15} = 2 x$

Step 2

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

${\sqrt{17 x - 15}}^{2} = {(2 x)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{17 x - 15}$ as ${(17 x - 15)}^{\frac{1}{2}}$ .

${({(17 x - 15)}^{\frac{1}{2}})}^{2} = {(2 x)}^{2}$

Simplify the left side.

Simplify ${({(17 x - 15)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(17 x - 15)}^{\frac{1}{2}})}^{2}$ .

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

${(17 x - 15)}^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(17 x - 15)}^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Rewrite the expression.

${(17 x - 15)}^{1} = {(2 x)}^{2}$

Simplify.

$17 x - 15 = {(2 x)}^{2}$

Simplify the right side.

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

Apply the product rule to $2 x$ .

$17 x - 15 = 2^{2} x^{2}$

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

$17 x - 15 = 4 x^{2}$

Step 4

Solve for $x$ .

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

$17 x - 15 - 4 x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $17 x - 15 - 4 x^{2}$ .

Reorder the expression.

Move $- 15$ .

$17 x - 4 x^{2} - 15 = 0$

Reorder $17 x$ and $- 4 x^{2}$ .

$- 4 x^{2} + 17 x - 15 = 0$

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

$- (4 x^{2}) + 17 x - 15 = 0$

Factor $- 1$ out of $17 x$ .

$- (4 x^{2}) - (- 17 x) - 15 = 0$

Rewrite $- 15$ as $- 1 (15)$ .

$- (4 x^{2}) - (- 17 x) - 1 \cdot 15 = 0$

Factor $- 1$ out of $- (4 x^{2}) - (- 17 x)$ .

$- (4 x^{2} - 17 x) - 1 \cdot 15 = 0$

Factor $- 1$ out of $- (4 x^{2} - 17 x) - 1 (15)$ .

$- (4 x^{2} - 17 x + 15) = 0$

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 $a \cdot c = 4 \cdot 15 = 60$ and whose sum is $b = - 17$ .

Factor $- 17$ out of $- 17 x$ .

$- (4 x^{2} - 17 x + 15) = 0$

Rewrite $- 17$ as $- 5$ plus $- 12$

$- (4 x^{2} + (- 5 - 12) x + 15) = 0$

Apply the distributive property.

$- (4 x^{2} - 5 x - 12 x + 15) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((4 x^{2} - 5 x) - 12 x + 15) = 0$

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

$- (x (4 x - 5) - 3 (4 x - 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 5$ .

$- ((4 x - 5) (x - 3)) = 0$

Remove unnecessary parentheses.

$- (4 x - 5) (x - 3) = 0$

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

$4 x - 5 = 0$

$x - 3 = 0$

Set $4 x - 5$ equal to $0$ and solve for $x$ .

Set $4 x - 5$ equal to $0$ .

$4 x - 5 = 0$

Solve $4 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$4 x = 5$

Divide each term in $4 x = 5$ by $4$ and simplify.

Divide each term in $4 x = 5$ by $4$ .

$\frac{4 x}{4} = \frac{5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{5}{4}$

Divide $x$ by $1$ .

$x = \frac{5}{4}$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $- (4 x - 5) (x - 3) = 0$ true.

$x = \frac{5}{4}, 3$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{4}, 3$

Decimal Form:

$x = 1.25, 3$

Mixed Number Form:

$x = 1 \frac{1}{4}, 3$

Solve for x 2x = square root of 17x-15

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