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What is the largest value of $x$ such that the expression

$$$\dfrac{x+1}{8x^2-65x+8}$$$

is not defined?

Guest Dec 29, 2017

#1
+5931
+1

The expression  $$\frac{x+1}{8x^2-65x+8}$$   is not defined when the denominator equals zero.

So to find all the  x  values that cause the expression to be undefined, set the denominator = 0 .

8x2 - 65x + 8  =  0            Now we need to solve this equation for  x .

We can factor the left side like this....

8x2 - 64x - x + 8  =  0

Factor  8x  out of the first two terms, factor  -1  out of the last two terms.

8x(x - 8) - 1(x - 8)  =  0

Factor  x-8  out of both terms.

(x - 8)(8x - 1)  =  0

Set each factor equal to zero and solve for  x .

x - 8  =  0          or          8x - 1  =  0

x  =  8               or          8x  =  1

x  =  1/8

So the expression is not defined when  x = 8  and when  x = 1/8  .

The largest of these values is  8 .

hectictar  Dec 29, 2017
Sort:

#1
+5931
+1

The expression  $$\frac{x+1}{8x^2-65x+8}$$   is not defined when the denominator equals zero.

So to find all the  x  values that cause the expression to be undefined, set the denominator = 0 .

8x2 - 65x + 8  =  0            Now we need to solve this equation for  x .

We can factor the left side like this....

8x2 - 64x - x + 8  =  0

Factor  8x  out of the first two terms, factor  -1  out of the last two terms.

8x(x - 8) - 1(x - 8)  =  0

Factor  x-8  out of both terms.

(x - 8)(8x - 1)  =  0

Set each factor equal to zero and solve for  x .

x - 8  =  0          or          8x - 1  =  0

x  =  8               or          8x  =  1

x  =  1/8

So the expression is not defined when  x = 8  and when  x = 1/8  .

The largest of these values is  8 .

hectictar  Dec 29, 2017

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