Solve for x x-3 square root of x-4=0

$x - 3 \sqrt{x} - 4 = 0$

Step 1

Move all terms not containing $- 3 \sqrt{x}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$- 3 \sqrt{x} - 4 = - x$

Add $4$ to both sides of the equation.

$- 3 \sqrt{x} = - x + 4$

Step 2

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

${(- 3 \sqrt{x})}^{2} = {(- x + 4)}^{2}$

Step 3

Simplify each side of the equation.

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

${(- 3 x^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Simplify the left side.

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

Apply the product rule to $- 3 x^{\frac{1}{2}}$ .

${(- 3)}^{2} {(x^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Raise $- 3$ to the power of $2$ .

$9 {(x^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

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

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

$9 x^{\frac{1}{2} \cdot 2} = {(- x + 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 x^{\frac{1}{2} \cdot 2} = {(- x + 4)}^{2}$

Rewrite the expression.

$9 x^{1} = {(- x + 4)}^{2}$

Simplify.

$9 x = {(- x + 4)}^{2}$

Simplify the right side.

Simplify ${(- x + 4)}^{2}$ .

Rewrite ${(- x + 4)}^{2}$ as $(- x + 4) (- x + 4)$ .

$9 x = (- x + 4) (- x + 4)$

Expand $(- x + 4) (- x + 4)$ using the FOIL Method.

Apply the distributive property.

$9 x = - x (- x + 4) + 4 (- x + 4)$

Apply the distributive property.

$9 x = - x (- x) - x \cdot 4 + 4 (- x + 4)$

Apply the distributive property.

$9 x = - x (- x) - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$9 x = - 1 \cdot - 1 x \cdot x - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$9 x = - 1 \cdot - 1 (x \cdot x) - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x$ by $x$ .

$9 x = - 1 \cdot - 1 x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $- 1$ by $- 1$ .

$9 x = 1 x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x^{2}$ by $1$ .

$9 x = x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $4$ by $- 1$ .

$9 x = x^{2} - 4 x + 4 (- x) + 4 \cdot 4$

Multiply $- 1$ by $4$ .

$9 x = x^{2} - 4 x - 4 x + 4 \cdot 4$

Multiply $4$ by $4$ .

$9 x = x^{2} - 4 x - 4 x + 16$

Subtract $4 x$ from $- 4 x$ .

$9 x = x^{2} - 8 x + 16$

Step 4

Solve for $x$ .

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

$x^{2} - 8 x + 16 = 9 x$

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

Subtract $9 x$ from both sides of the equation.

$x^{2} - 8 x + 16 - 9 x = 0$

Subtract $9 x$ from $- 8 x$ .

$x^{2} - 17 x + 16 = 0$

Factor $x^{2} - 17 x + 16$ 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 $16$ and whose sum is $- 17$ .

$- 16, - 1$

Write the factored form using these integers.

$(x - 16) (x - 1) = 0$

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

$x - 16 = 0$

$x - 1 = 0$

Set $x - 16$ equal to $0$ and solve for $x$ .

Set $x - 16$ equal to $0$ .

$x - 16 = 0$

Add $16$ to both sides of the equation.

$x = 16$

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.

$x = 1$

The final solution is all the values that make $(x - 16) (x - 1) = 0$ true.

$x = 16, 1$

Step 5

Exclude the solutions that do not make $x - 3 \sqrt{x} - 4 = 0$ true.

$x = 16$

Solve for x x-3 square root of x-4=0

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