Solve for x square root of 6-5x=x

$\sqrt{6 - 5 x} = x$

Step 1

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

${\sqrt{6 - 5 x}}^{2} = x^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6 - 5 x}$ as ${(6 - 5 x)}^{\frac{1}{2}}$ .

${({(6 - 5 x)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(6 - 5 x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(6 - 5 x)}^{\frac{1}{2}})}^{2}$ .

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

${(6 - 5 x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(6 - 5 x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(6 - 5 x)}^{1} = x^{2}$

Simplify.

$6 - 5 x = x^{2}$

Step 3

Solve for $x$ .

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

$6 - 5 x - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $6 - 5 x - x^{2}$ .

Reorder the expression.

Move $6$ .

$- 5 x - x^{2} + 6 = 0$

Reorder $- 5 x$ and $- x^{2}$ .

$- x^{2} - 5 x + 6 = 0$

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

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

Factor $- 1$ out of $- 5 x$ .

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

Rewrite $6$ as $- 1 (- 6)$ .

$- (x^{2}) - (5 x) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (x^{2}) - (5 x)$ .

$- (x^{2} + 5 x) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (x^{2} + 5 x) - 1 (- 6)$ .

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

Factor.

Factor $x^{2} + 5 x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $5$ .

$- 1, 6$

Write the factored form using these integers.

$- ((x - 1) (x + 6)) = 0$

Remove unnecessary parentheses.

$- (x - 1) (x + 6) = 0$

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

$x - 1 = 0$

$x + 6 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- (x - 1) (x + 6) = 0$ true.

$x = 1, - 6$

Step 4

Exclude the solutions that do not make $\sqrt{6 - 5 x} = x$ true.

$x = 1$

Solve for x square root of 6-5x=x

Related Questions

Support LCMGCF.com by disabling your adblocker.