Question Solve the equation, eliminating any extraneous solutions. $\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}$ . The solutions are $x=-5,5$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}$ .
2. We need to solve for $x$ .
3. Extraneous solutions may arise when we manipulate the equation, and we must check for them at the end.

STEP 2

Since the denominators on both sides of the equation are the same, we can equate the numerators as long as $x+5 \neq 0$ (to avoid division by zero).
$x^{2}+5 = 30$

STEP 3

Subtract 5 from both sides of the equation to isolate the quadratic term.
$x^{2}+5 - 5 = 30 - 5$

STEP 4

Simplify the equation.
$x^{2} = 25$

STEP 5

Take the square root of both sides of the equation to solve for $x$ . Remember that taking the square root gives us both positive and negative solutions.
$x = \pm\sqrt{25}$

STEP 6

Calculate the square root of 25.
$x = \pm5$

STEP 7

We have two potential solutions, $x = 5$ and $x = -5$ . However, we must check for extraneous solutions by substituting these values back into the original equation.

STEP 8

First, check $x = 5$ by substituting it into the original equation.
$\frac{5^{2}+5}{(5+5)} \stackrel{?}{=} \frac{30}{(5+5)}$

STEP 9

Simplify the left side of the equation.
$\frac{25+5}{10} = \frac{30}{10}$

STEP 10

Further simplify the left side of the equation.
$\frac{30}{10} = \frac{30}{10}$

STEP 11

Since the equation is true for $x = 5$ , it is not an extraneous solution.

STEP 12

Now, check $x = -5$ by substituting it into the original equation.
$\frac{(-5)^{2}+5}{(-5+5)} \stackrel{?}{=} \frac{30}{(-5+5)}$

STEP 13

Simplify the left side of the equation, noting that the denominator becomes zero.
$\frac{25+5}{0} \stackrel{?}{=} \frac{30}{0}$

STEP 14

Since division by zero is undefined, $x = -5$ creates an undefined expression on both sides of the equation. Therefore, $x = -5$ is an extraneous solution and cannot be a valid solution to the original equation.

STEP 15

The only valid solution to the original equation is $x = 5$ .
$x=5$

Was this helpful?

Question Solve the equation, eliminating any extraneous solutions. $\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}$ . The solutions are $x=-5,5$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}$ .
2. We need to solve for $x$ .
3. Extraneous solutions may arise when we manipulate the equation, and we must check for them at the end.

STEP 2

Since the denominators on both sides of the equation are the same, we can equate the numerators as long as $x+5 \neq 0$ (to avoid division by zero).
$x^{2}+5 = 30$

STEP 3

Subtract 5 from both sides of the equation to isolate the quadratic term.
$x^{2}+5 - 5 = 30 - 5$

STEP 4

Simplify the equation.
$x^{2} = 25$

STEP 5

Take the square root of both sides of the equation to solve for $x$ . Remember that taking the square root gives us both positive and negative solutions.
$x = \pm\sqrt{25}$

STEP 6

Calculate the square root of 25.
$x = \pm5$

STEP 7

We have two potential solutions, $x = 5$ and $x = -5$ . However, we must check for extraneous solutions by substituting these values back into the original equation.

STEP 8

First, check $x = 5$ by substituting it into the original equation.
$\frac{5^{2}+5}{(5+5)} \stackrel{?}{=} \frac{30}{(5+5)}$

STEP 9

Simplify the left side of the equation.
$\frac{25+5}{10} = \frac{30}{10}$

STEP 10

Further simplify the left side of the equation.
$\frac{30}{10} = \frac{30}{10}$

STEP 11

Since the equation is true for $x = 5$ , it is not an extraneous solution.

STEP 12

Now, check $x = -5$ by substituting it into the original equation.
$\frac{(-5)^{2}+5}{(-5+5)} \stackrel{?}{=} \frac{30}{(-5+5)}$

STEP 13

Simplify the left side of the equation, noting that the denominator becomes zero.
$\frac{25+5}{0} \stackrel{?}{=} \frac{30}{0}$

STEP 14

Since division by zero is undefined, $x = -5$ creates an undefined expression on both sides of the equation. Therefore, $x = -5$ is an extraneous solution and cannot be a valid solution to the original equation.

STEP 15

The only valid solution to the original equation is $x = 5$ .
$x=5$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question Solve the equation, eliminating any extraneous solutions. x2+5(x+5)=30(x+5)\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}(x+5)x2+5​=(x+5)30​. The solutions are x=−5,5x=-5,5x=−5,5.

Studdy Solution

STEP 1

Assumptions1. We are given the equation x2+5(x+5)=30(x+5)\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}(x+5)x2+5​=(x+5)30​.2. We need to solve for xxx.3. Extraneous solutions may arise when we manipulate the equation, and we must check for them at the end.

STEP 2

Since the denominators on both sides of the equation are the same, we can equate the numerators as long as x+5≠0x+5 \neq 0x+5=0 (to avoid division by zero).x2+5=30x^{2}+5 = 30x2+5=30

STEP 3

Subtract 5 from both sides of the equation to isolate the quadratic term.x2+5−5=30−5x^{2}+5 - 5 = 30 - 5x2+5−5=30−5

STEP 4

Simplify the equation.x2=25x^{2} = 25x2=25

STEP 5

Take the square root of both sides of the equation to solve for xxx. Remember that taking the square root gives us both positive and negative solutions.x=±25x = \pm\sqrt{25}x=±25​

STEP 6

Calculate the square root of 25.x=±5x = \pm5x=±5

STEP 7

We have two potential solutions, x=5x = 5x=5 and x=−5x = -5x=−5. However, we must check for extraneous solutions by substituting these values back into the original equation.

STEP 8

First, check x=5x = 5x=5 by substituting it into the original equation.52+5(5+5)=?30(5+5)\frac{5^{2}+5}{(5+5)} \stackrel{?}{=} \frac{30}{(5+5)}(5+5)52+5​=?(5+5)30​

STEP 9

Simplify the left side of the equation.25+510=3010\frac{25+5}{10} = \frac{30}{10}1025+5​=1030​

STEP 10

Further simplify the left side of the equation.3010=3010\frac{30}{10} = \frac{30}{10}1030​=1030​

STEP 11

Since the equation is true for x=5x = 5x=5, it is not an extraneous solution.

STEP 12

Now, check x=−5x = -5x=−5 by substituting it into the original equation.(−5)2+5(−5+5)=?30(−5+5)\frac{(-5)^{2}+5}{(-5+5)} \stackrel{?}{=} \frac{30}{(-5+5)}(−5+5)(−5)2+5​=?(−5+5)30​

STEP 13

Simplify the left side of the equation, noting that the denominator becomes zero.25+50=?300\frac{25+5}{0} \stackrel{?}{=} \frac{30}{0}025+5​=?030​

STEP 14

Since division by zero is undefined, x=−5x = -5x=−5 creates an undefined expression on both sides of the equation. Therefore, x=−5x = -5x=−5 is an extraneous solution and cannot be a valid solution to the original equation.

STEP 15

The only valid solution to the original equation is x=5x = 5x=5.x=5x=5x=5

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question Solve the equation, eliminating any extraneous solutions. $\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}$ . The solutions are $x=-5,5$ .

Assumptions
1. We are given the equation $\frac{x^{2}+5}{(x+5)}=\frac{30}{(x+5)}$ .
2. We need to solve for $x$ .
3. Extraneous solutions may arise when we manipulate the equation, and we must check for them at the end.

Since the denominators on both sides of the equation are the same, we can equate the numerators as long as $x+5 \neq 0$ (to avoid division by zero).
$x^{2}+5 = 30$

Subtract 5 from both sides of the equation to isolate the quadratic term.
$x^{2}+5 - 5 = 30 - 5$

Simplify the equation.
$x^{2} = 25$

Take the square root of both sides of the equation to solve for $x$ . Remember that taking the square root gives us both positive and negative solutions.
$x = \pm\sqrt{25}$

Calculate the square root of 25.
$x = \pm5$

We have two potential solutions, $x = 5$ and $x = -5$ . However, we must check for extraneous solutions by substituting these values back into the original equation.

First, check $x = 5$ by substituting it into the original equation.
$\frac{5^{2}+5}{(5+5)} \stackrel{?}{=} \frac{30}{(5+5)}$

Simplify the left side of the equation.
$\frac{25+5}{10} = \frac{30}{10}$

Further simplify the left side of the equation.
$\frac{30}{10} = \frac{30}{10}$

Since the equation is true for $x = 5$ , it is not an extraneous solution.

Now, check $x = -5$ by substituting it into the original equation.
$\frac{(-5)^{2}+5}{(-5+5)} \stackrel{?}{=} \frac{30}{(-5+5)}$

Simplify the left side of the equation, noting that the denominator becomes zero.
$\frac{25+5}{0} \stackrel{?}{=} \frac{30}{0}$

Since division by zero is undefined, $x = -5$ creates an undefined expression on both sides of the equation. Therefore, $x = -5$ is an extraneous solution and cannot be a valid solution to the original equation.

The only valid solution to the original equation is $x = 5$ .
$x=5$