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zzz [600]
3 years ago
7

Given the equation Square root of 8x plus 1 = 5, solve for x and identify if it is an extraneous solution.

Mathematics
2 answers:
Ksivusya [100]3 years ago
7 0

For this case we have the following equation:

\sqrt{8x+1}=5

From here, we must clear the value of x.

For this, we follow the following steps:

1) We raise both members of the equation to the square:

8x+1=25

2) We pass the number 1 subtracting:

8x = 25 - 1  8x = 24

3) We pass number 8 dividing :

x =\frac{24}{8} = 3

We observe that the given equation can be evaluated for the solution obtained.

Therefore, the solution is valid.

Answer:

C) x = 3, solution is not extraneous

Rasek [7]3 years ago
6 0
An extraneous solution is a root that when substituted to the original equation, the answer becomes invalid. In this respect, the equation is square root (8x + 1) = 5; 8x + 1 = 25; x1 = 3;  <span>8x + 1 = -25 ; x2 = -26/8 = -13 /4. Upon substitution, square root of (8*3 + 1 ) = 5, this is valid. Hence the answer to this question is C.</span>
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2 years ago
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Answer:

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