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The problem: take two solutions and plug them into the original equation. To check for extraneous solutions.

The problem: take two solutions and plug them into the original equation. To check-example-1
User Docuemada
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1 Answer

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Answer: х=3.

Explanation:

А point is missing here: determining the allowable values of x for this equation.

1. Consider the right side of the equation:

The root expression 2x+3 must be greater than or equal to 0.

2x+3≥0

2x≥-3\ |:2

x≥-1,5.

2. Consider the left side of the equation:

the left side of the equation must be greater than or equal to 0.

x≥0.

We get x≥0. ⇒

х=3 ∈ х≥0

х=-1 ∉ х≥0.

User SyntaxError
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