Solved on Oct 25, 2023

Find real solutions to the equation $\frac{5}{x^{2}-5x}-\frac{1}{x^{2}-25}=0$ .

STEP 1

Assumptions1. We are looking for all real solutions. . The given equation is $\frac{5}{x^{}-5 x}-\frac{1}{x^{}-25}=0$ .

STEP 2

First, we need to simplify the equation. We can do this by finding a common denominator for the two fractions.
$\frac{5}{x^{2}-5 x}-\frac{1}{x^{2}-25} = \frac{5(x^{2}-25) - (x^{2}-5x)}{(x^{2}-5x)(x^{2}-25)}$

STEP 3

Now, expand the numerator and simplify the equation.
$\frac{5x^{2}-125 - x^{2}+5x}{(x^{2}-5x)(x^{2}-25)} = \frac{x^{2}+5x-125}{(x^{2}-5x)(x^{2}-25)}$

STEP 4

The equation $\frac{4x^{2}+x-125}{(x^{2}-x)(x^{2}-25)} =0$ holds true if and only if the numerator is zero. So we solve the equation $4x^{2}+x-125 =0$ .

STEP 5

This is a quadratic equation, and we can solve it using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ .

STEP 6

Plug in the values for a, b, and c from the equation $4x^{2}+5x-125 =0$ into the quadratic formula.
$x = \frac{-5 \pm \sqrt{5^{2}-44(-125)}}{2*4}$

STEP 7

implify the equation under the square root.
$x = \frac{-5 \pm \sqrt{25+2000}}{}$

STEP 8

Calculate the value under the square root.
$x = \frac{-5 \pm \sqrt{2025}}{8}$

STEP 9

Take the square root of2025.
$x = \frac{-5 \pm45}{8}$

STEP 10

Calculate the two possible solutions.
$x = \frac{-5 +45}{8} =5, \quad x = \frac{-5 -45}{8} = -6.25$

STEP 11

Now we need to check these solutions in the original equation to ensure they are valid.For $x =5$ , the original equation becomes undefined, so $x =5$ is not a valid solution.
For $x = -6.25$ , substitute this value into the original equation and check if it holds true.

STEP 12

Substitute $x = -6.25$ into the original equation.
$\frac{5}{(-6.25)^{2}-5*(-6.25)}-\frac{}{(-6.25)^{2}-25} =0$

STEP 13

Calculate the values in the equation.
$\frac{5}{39.0625+31.25}-\frac{}{39.0625-25} =0$

STEP 14

Calculate the values in the equation.
$\frac{}{70.312}-\frac{}{14.062} =0$

STEP 15

Calculate the values in the equation.
$0.071 -0.071 =0$

STEP 16

The equation holds true, so $x = -6.25$ is a valid solution.
The only real solution to the equation $\frac{5}{x^{2}-5 x}-\frac{}{x^{2}-25}=0$ is $x = -6.25$ .

Was this helpful?

Find real solutions to the equation $\frac{5}{x^{2}-5x}-\frac{1}{x^{2}-25}=0$ .

STEP 1

Assumptions1. We are looking for all real solutions. . The given equation is $\frac{5}{x^{}-5 x}-\frac{1}{x^{}-25}=0$ .

STEP 2

First, we need to simplify the equation. We can do this by finding a common denominator for the two fractions.
$\frac{5}{x^{2}-5 x}-\frac{1}{x^{2}-25} = \frac{5(x^{2}-25) - (x^{2}-5x)}{(x^{2}-5x)(x^{2}-25)}$

STEP 3

Now, expand the numerator and simplify the equation.
$\frac{5x^{2}-125 - x^{2}+5x}{(x^{2}-5x)(x^{2}-25)} = \frac{x^{2}+5x-125}{(x^{2}-5x)(x^{2}-25)}$

STEP 4

The equation $\frac{4x^{2}+x-125}{(x^{2}-x)(x^{2}-25)} =0$ holds true if and only if the numerator is zero. So we solve the equation $4x^{2}+x-125 =0$ .

STEP 5

This is a quadratic equation, and we can solve it using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ .

STEP 6

Plug in the values for a, b, and c from the equation $4x^{2}+5x-125 =0$ into the quadratic formula.
$x = \frac{-5 \pm \sqrt{5^{2}-44(-125)}}{2*4}$

STEP 7

implify the equation under the square root.
$x = \frac{-5 \pm \sqrt{25+2000}}{}$

STEP 8

Calculate the value under the square root.
$x = \frac{-5 \pm \sqrt{2025}}{8}$

STEP 9

Take the square root of2025.
$x = \frac{-5 \pm45}{8}$

STEP 10

Calculate the two possible solutions.
$x = \frac{-5 +45}{8} =5, \quad x = \frac{-5 -45}{8} = -6.25$

STEP 11

Now we need to check these solutions in the original equation to ensure they are valid.For $x =5$ , the original equation becomes undefined, so $x =5$ is not a valid solution.
For $x = -6.25$ , substitute this value into the original equation and check if it holds true.

STEP 12

Substitute $x = -6.25$ into the original equation.
$\frac{5}{(-6.25)^{2}-5*(-6.25)}-\frac{}{(-6.25)^{2}-25} =0$

STEP 13

Calculate the values in the equation.
$\frac{5}{39.0625+31.25}-\frac{}{39.0625-25} =0$

STEP 14

Calculate the values in the equation.
$\frac{}{70.312}-\frac{}{14.062} =0$

STEP 15

Calculate the values in the equation.
$0.071 -0.071 =0$

STEP 16

The equation holds true, so $x = -6.25$ is a valid solution.
The only real solution to the equation $\frac{5}{x^{2}-5 x}-\frac{}{x^{2}-25}=0$ is $x = -6.25$ .

Start learning now

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

Find real solutions to the equation 5x2−5x−1x2−25=0\frac{5}{x^{2}-5x}-\frac{1}{x^{2}-25}=0x2−5x5​−x2−251​=0.

STEP 1

Assumptions1. We are looking for all real solutions. . The given equation is 5x−5x−1x−25=0\frac{5}{x^{}-5 x}-\frac{1}{x^{}-25}=0x−5x5​−x−251​=0.

STEP 2

STEP 3

Now, expand the numerator and simplify the equation.5x2−125−x2+5x(x2−5x)(x2−25)=x2+5x−125(x2−5x)(x2−25)\frac{5x^{2}-125 - x^{2}+5x}{(x^{2}-5x)(x^{2}-25)} = \frac{x^{2}+5x-125}{(x^{2}-5x)(x^{2}-25)}(x2−5x)(x2−25)5x2−125−x2+5x​=(x2−5x)(x2−25)x2+5x−125​

STEP 4

The equation 4x2+x−125(x2−x)(x2−25)=0\frac{4x^{2}+x-125}{(x^{2}-x)(x^{2}-25)} =0(x2−x)(x2−25)4x2+x−125​=0 holds true if and only if the numerator is zero. So we solve the equation 4x2+x−125=04x^{2}+x-125 =04x2+x−125=0.

STEP 5

This is a quadratic equation, and we can solve it using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}x=2a−b±b2−4ac​​.

STEP 6

Plug in the values for a, b, and c from the equation 4x2+5x−125=04x^{2}+5x-125 =04x2+5x−125=0 into the quadratic formula.x=−5±52−4∗4∗(−125)2∗4x = \frac{-5 \pm \sqrt{5^{2}-4*4*(-125)}}{2*4}x=2∗4−5±52−4∗4∗(−125)​​

STEP 7

implify the equation under the square root.x=−5±25+2000x = \frac{-5 \pm \sqrt{25+2000}}{}x=−5±25+2000​​

STEP 8

Calculate the value under the square root.x=−5±20258x = \frac{-5 \pm \sqrt{2025}}{8}x=8−5±2025​​

STEP 9

Take the square root of2025.x=−5±458x = \frac{-5 \pm45}{8}x=8−5±45​

STEP 10

Calculate the two possible solutions.x=−5+458=5,x=−5−458=−6.25x = \frac{-5 +45}{8} =5, \quad x = \frac{-5 -45}{8} = -6.25x=8−5+45​=5,x=8−5−45​=−6.25

STEP 11

Now we need to check these solutions in the original equation to ensure they are valid.For x=5x =5x=5, the original equation becomes undefined, so x=5x =5x=5 is not a valid solution.For x=−6.25x = -6.25x=−6.25, substitute this value into the original equation and check if it holds true.

STEP 12

Substitute x=−6.25x = -6.25x=−6.25 into the original equation.5(−6.25)2−5∗(−6.25)−(−6.25)2−25=0\frac{5}{(-6.25)^{2}-5*(-6.25)}-\frac{}{(-6.25)^{2}-25} =0(−6.25)2−5∗(−6.25)5​−(−6.25)2−25​=0

STEP 13

Calculate the values in the equation.539.0625+31.25−39.0625−25=0\frac{5}{39.0625+31.25}-\frac{}{39.0625-25} =039.0625+31.255​−39.0625−25​=0

STEP 14

Calculate the values in the equation.70.312−14.062=0\frac{}{70.312}-\frac{}{14.062} =070.312​−14.062​=0

STEP 15

Calculate the values in the equation.0.071−0.071=00.071 -0.071 =00.071−0.071=0

STEP 16

The equation holds true, so x=−6.25x = -6.25x=−6.25 is a valid solution.The only real solution to the equation 5x2−5x−x2−25=0\frac{5}{x^{2}-5 x}-\frac{}{x^{2}-25}=0x2−5x5​−x2−25​=0 is x=−6.25x = -6.25x=−6.25.

Start learning now

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

Find real solutions to the equation $\frac{5}{x^{2}-5x}-\frac{1}{x^{2}-25}=0$ .

Assumptions1. We are looking for all real solutions. . The given equation is $\frac{5}{x^{}-5 x}-\frac{1}{x^{}-25}=0$ .

Now, expand the numerator and simplify the equation.
$\frac{5x^{2}-125 - x^{2}+5x}{(x^{2}-5x)(x^{2}-25)} = \frac{x^{2}+5x-125}{(x^{2}-5x)(x^{2}-25)}$

The equation $\frac{4x^{2}+x-125}{(x^{2}-x)(x^{2}-25)} =0$ holds true if and only if the numerator is zero. So we solve the equation $4x^{2}+x-125 =0$ .

This is a quadratic equation, and we can solve it using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ .

Plug in the values for a, b, and c from the equation $4x^{2}+5x-125 =0$ into the quadratic formula.
$x = \frac{-5 \pm \sqrt{5^{2}-44(-125)}}{2*4}$

implify the equation under the square root.
$x = \frac{-5 \pm \sqrt{25+2000}}{}$

Calculate the value under the square root.
$x = \frac{-5 \pm \sqrt{2025}}{8}$

Take the square root of2025.
$x = \frac{-5 \pm45}{8}$

Calculate the two possible solutions.
$x = \frac{-5 +45}{8} =5, \quad x = \frac{-5 -45}{8} = -6.25$

Now we need to check these solutions in the original equation to ensure they are valid.For $x =5$ , the original equation becomes undefined, so $x =5$ is not a valid solution.
For $x = -6.25$ , substitute this value into the original equation and check if it holds true.

Substitute $x = -6.25$ into the original equation.
$\frac{5}{(-6.25)^{2}-5*(-6.25)}-\frac{}{(-6.25)^{2}-25} =0$

Calculate the values in the equation.
$\frac{5}{39.0625+31.25}-\frac{}{39.0625-25} =0$

Calculate the values in the equation.
$\frac{}{70.312}-\frac{}{14.062} =0$

Calculate the values in the equation.
$0.071 -0.071 =0$

The equation holds true, so $x = -6.25$ is a valid solution.
The only real solution to the equation $\frac{5}{x^{2}-5 x}-\frac{}{x^{2}-25}=0$ is $x = -6.25$ .