Solve by Factoring square root of 21x+1=x+5

$\sqrt{21 x + 1} = x + 5$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the equation.

$\sqrt{21 x + 1} - x = 5$

Subtract $5$ from both sides of the equation.

$\sqrt{21 x + 1} - x - 5 = 0$

Step 2

Move all terms not containing $\sqrt{21 x + 1}$ to the right side of the equation.

Add $x$ to both sides of the equation.

$\sqrt{21 x + 1} - 5 = x$

Add $5$ to both sides of the equation.

$\sqrt{21 x + 1} = x + 5$

Step 3

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

${\sqrt{21 x + 1}}^{2} = {(x + 5)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{21 x + 1}$ as ${(21 x + 1)}^{\frac{1}{2}}$ .

${({(21 x + 1)}^{\frac{1}{2}})}^{2} = {(x + 5)}^{2}$

Simplify the left side.

Simplify ${({(21 x + 1)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(21 x + 1)}^{\frac{1}{2}})}^{2}$ .

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

${(21 x + 1)}^{\frac{1}{2} \cdot 2} = {(x + 5)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(21 x + 1)}^{\frac{1}{2} \cdot 2} = {(x + 5)}^{2}$

Rewrite the expression.

${(21 x + 1)}^{1} = {(x + 5)}^{2}$

Simplify.

$21 x + 1 = {(x + 5)}^{2}$

Simplify the right side.

Simplify ${(x + 5)}^{2}$ .

Rewrite ${(x + 5)}^{2}$ as $(x + 5) (x + 5)$ .

$21 x + 1 = (x + 5) (x + 5)$

Expand $(x + 5) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$21 x + 1 = x (x + 5) + 5 (x + 5)$

Apply the distributive property.

$21 x + 1 = x \cdot x + x \cdot 5 + 5 (x + 5)$

Apply the distributive property.

$21 x + 1 = x \cdot x + x \cdot 5 + 5 x + 5 \cdot 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$21 x + 1 = x^{2} + x \cdot 5 + 5 x + 5 \cdot 5$

Move $5$ to the left of $x$ .

$21 x + 1 = x^{2} + 5 \cdot x + 5 x + 5 \cdot 5$

Multiply $5$ by $5$ .

$21 x + 1 = x^{2} + 5 x + 5 x + 25$

Add $5 x$ and $5 x$ .

$21 x + 1 = x^{2} + 10 x + 25$

Step 5

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.

$x^{2} + 10 x + 25 = 21 x + 1$

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

Subtract $21 x$ from both sides of the equation.

$x^{2} + 10 x + 25 - 21 x = 1$

Subtract $21 x$ from $10 x$ .

$x^{2} - 11 x + 25 = 1$

Subtract $1$ from both sides of the equation.

$x^{2} - 11 x + 25 - 1 = 0$

Subtract $1$ from $25$ .

$x^{2} - 11 x + 24 = 0$

Factor $x^{2} - 11 x + 24$ 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 $24$ and whose sum is $- 11$ .

$- 8, - 3$

Write the factored form using these integers.

$(x - 8) (x - 3) = 0$

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

$x - 8 = 0$

$x - 3 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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 $(x - 8) (x - 3) = 0$ true.

$x = 8, 3$

Solve by Factoring square root of 21x+1=x+5

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