Solve for x 5 square root of x-1=x+3

$5 \sqrt{x - 1} = x + 3$

Step 1

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

${(5 \sqrt{x - 1})}^{2} = {(x + 3)}^{2}$

Step 2

Simplify each side of the equation.

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

${(5 {(x - 1)}^{\frac{1}{2}})}^{2} = {(x + 3)}^{2}$

Simplify the left side.

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

Apply the product rule to $5 {(x - 1)}^{\frac{1}{2}}$ .

$5^{2} {({(x - 1)}^{\frac{1}{2}})}^{2} = {(x + 3)}^{2}$

Raise $5$ to the power of $2$ .

$25 {({(x - 1)}^{\frac{1}{2}})}^{2} = {(x + 3)}^{2}$

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

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

$25 {(x - 1)}^{\frac{1}{2} \cdot 2} = {(x + 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$25 {(x - 1)}^{\frac{1}{2} \cdot 2} = {(x + 3)}^{2}$

Rewrite the expression.

$25 {(x - 1)}^{1} = {(x + 3)}^{2}$

Simplify.

$25 (x - 1) = {(x + 3)}^{2}$

Apply the distributive property.

$25 x + 25 \cdot - 1 = {(x + 3)}^{2}$

Multiply $25$ by $- 1$ .

$25 x - 25 = {(x + 3)}^{2}$

Simplify the right side.

Simplify ${(x + 3)}^{2}$ .

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

$25 x - 25 = (x + 3) (x + 3)$

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$25 x - 25 = x (x + 3) + 3 (x + 3)$

Apply the distributive property.

$25 x - 25 = x \cdot x + x \cdot 3 + 3 (x + 3)$

Apply the distributive property.

$25 x - 25 = x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$25 x - 25 = x^{2} + x \cdot 3 + 3 x + 3 \cdot 3$

Move $3$ to the left of $x$ .

$25 x - 25 = x^{2} + 3 \cdot x + 3 x + 3 \cdot 3$

Multiply $3$ by $3$ .

$25 x - 25 = x^{2} + 3 x + 3 x + 9$

Add $3 x$ and $3 x$ .

$25 x - 25 = x^{2} + 6 x + 9$

Step 3

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.

$x^{2} + 6 x + 9 = 25 x - 25$

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

Subtract $25 x$ from both sides of the equation.

$x^{2} + 6 x + 9 - 25 x = - 25$

Subtract $25 x$ from $6 x$ .

$x^{2} - 19 x + 9 = - 25$

Add $25$ to both sides of the equation.

$x^{2} - 19 x + 9 + 25 = 0$

Add $9$ and $25$ .

$x^{2} - 19 x + 34 = 0$

Factor $x^{2} - 19 x + 34$ 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 $34$ and whose sum is $- 19$ .

$- 17, - 2$

Write the factored form using these integers.

$(x - 17) (x - 2) = 0$

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

$x - 17 = 0$

$x - 2 = 0$

Set $x - 17$ equal to $0$ and solve for $x$ .

Set $x - 17$ equal to $0$ .

$x - 17 = 0$

Add $17$ to both sides of the equation.

$x = 17$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x - 17) (x - 2) = 0$ true.

$x = 17, 2$

Solve for x 5 square root of x-1=x+3

Related Questions

Support LCMGCF.com by disabling your adblocker.