Solved on Sep 19, 2023

Find the values of $x$ for which the equation $\frac{8}{(x-2)^{2}}=\frac{4}{x^{2}-9}+\frac{2}{3}$ is undefined.

STEP 1

Assumptions1. The given equation is $\frac{8}{(x-)^{}}=\frac{4}{x^{}-9}+\frac{}{3}$ . . We are asked to find the excluded values, which are the values of $x$ that would make the denominator of any fraction in the equation equal to zero.

STEP 2

First, let's find the values of $x$ that would make the denominator of the first fraction equal to zero. We set the denominator equal to zero and solve for $x$ .
$(x-2)^{2}=0$

STEP 3

olve the equation $(x-2)^{2}=0$ for $x$ .
$x-2=0$ $x=2$ So, $x=2$ is an excluded value because it would make the denominator of the first fraction equal to zero.

STEP 4

Next, let's find the values of $x$ that would make the denominator of the second fraction equal to zero. We set the denominator equal to zero and solve for $x$ .
$x^{2}-9=0$

STEP 5

olve the equation $x^{2}-9=0$ for $x$ .
$x^{2}=9$ $x=\pm\sqrt{9}$ $x=\pm3$ So, $x=-3$ and $x=3$ are excluded values because they would make the denominator of the second fraction equal to zero.

STEP 6

The third fraction does not have $x$ in its denominator, so it does not contribute any excluded values.
Therefore, the excluded values for the given equation are $x=2$ , $x=-3$ , and $x=3$ .

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Find the values of $x$ for which the equation $\frac{8}{(x-2)^{2}}=\frac{4}{x^{2}-9}+\frac{2}{3}$ is undefined.

STEP 1

Assumptions1. The given equation is $\frac{8}{(x-)^{}}=\frac{4}{x^{}-9}+\frac{}{3}$ . . We are asked to find the excluded values, which are the values of $x$ that would make the denominator of any fraction in the equation equal to zero.

STEP 2

First, let's find the values of $x$ that would make the denominator of the first fraction equal to zero. We set the denominator equal to zero and solve for $x$ .
$(x-2)^{2}=0$

STEP 3

olve the equation $(x-2)^{2}=0$ for $x$ .
$x-2=0$ $x=2$ So, $x=2$ is an excluded value because it would make the denominator of the first fraction equal to zero.

STEP 4

Next, let's find the values of $x$ that would make the denominator of the second fraction equal to zero. We set the denominator equal to zero and solve for $x$ .
$x^{2}-9=0$

STEP 5

olve the equation $x^{2}-9=0$ for $x$ .
$x^{2}=9$ $x=\pm\sqrt{9}$ $x=\pm3$ So, $x=-3$ and $x=3$ are excluded values because they would make the denominator of the second fraction equal to zero.

STEP 6

The third fraction does not have $x$ in its denominator, so it does not contribute any excluded values.
Therefore, the excluded values for the given equation are $x=2$ , $x=-3$ , and $x=3$ .

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Find the values of xxx for which the equation 8(x−2)2=4x2−9+23\frac{8}{(x-2)^{2}}=\frac{4}{x^{2}-9}+\frac{2}{3}(x−2)28​=x2−94​+32​ is undefined.

STEP 1

Assumptions1. The given equation is 8(x−)=4x−9+3\frac{8}{(x-)^{}}=\frac{4}{x^{}-9}+\frac{}{3}(x−)8​=x−94​+3​. . We are asked to find the excluded values, which are the values of xxx that would make the denominator of any fraction in the equation equal to zero.

STEP 2

First, let's find the values of xxx that would make the denominator of the first fraction equal to zero. We set the denominator equal to zero and solve for xxx.(x−2)2=0(x-2)^{2}=0(x−2)2=0

STEP 3

olve the equation (x−2)2=0(x-2)^{2}=0(x−2)2=0 for xxx.x−2=0x-2=0x−2=0x=2x=2x=2So, x=2x=2x=2 is an excluded value because it would make the denominator of the first fraction equal to zero.

STEP 4

Next, let's find the values of xxx that would make the denominator of the second fraction equal to zero. We set the denominator equal to zero and solve for xxx.x2−9=0x^{2}-9=0x2−9=0

STEP 5

olve the equation x2−9=0x^{2}-9=0x2−9=0 for xxx.x2=9x^{2}=9x2=9x=±9x=\pm\sqrt{9}x=±9​x=±3x=\pm3x=±3So, x=−3x=-3x=−3 and x=3x=3x=3 are excluded values because they would make the denominator of the second fraction equal to zero.

STEP 6

The third fraction does not have xxx in its denominator, so it does not contribute any excluded values.Therefore, the excluded values for the given equation are x=2x=2x=2, x=−3x=-3x=−3, and x=3x=3x=3.

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Find the values of $x$ for which the equation $\frac{8}{(x-2)^{2}}=\frac{4}{x^{2}-9}+\frac{2}{3}$ is undefined.

Assumptions1. The given equation is $\frac{8}{(x-)^{}}=\frac{4}{x^{}-9}+\frac{}{3}$ . . We are asked to find the excluded values, which are the values of $x$ that would make the denominator of any fraction in the equation equal to zero.

First, let's find the values of $x$ that would make the denominator of the first fraction equal to zero. We set the denominator equal to zero and solve for $x$ .
$(x-2)^{2}=0$

olve the equation $(x-2)^{2}=0$ for $x$ .
$x-2=0$ $x=2$ So, $x=2$ is an excluded value because it would make the denominator of the first fraction equal to zero.

Next, let's find the values of $x$ that would make the denominator of the second fraction equal to zero. We set the denominator equal to zero and solve for $x$ .
$x^{2}-9=0$

olve the equation $x^{2}-9=0$ for $x$ .
$x^{2}=9$ $x=\pm\sqrt{9}$ $x=\pm3$ So, $x=-3$ and $x=3$ are excluded values because they would make the denominator of the second fraction equal to zero.

The third fraction does not have $x$ in its denominator, so it does not contribute any excluded values.
Therefore, the excluded values for the given equation are $x=2$ , $x=-3$ , and $x=3$ .