Solve: $\quad \sqrt{-4 x-6}=-2$
Solution: $x=$ $\square$
Answer (2)
Square both sides of the equation − 4 x − 6 = − 2 to get − 4 x − 6 = 4 .
Solve for x , obtaining x = − 2 5 .
Substitute x = − 2 5 back into the original equation to check for extraneous solutions.
Since the substitution leads to 2 = − 2 , which is false, there is no real solution. The answer is D NE .
Explanation
Understanding the Problem We are given the equation − 4 x − 6 = − 2 . We need to solve for x . We must check for extraneous solutions, since we are dealing with a square root.
Squaring Both Sides To solve the equation, we first square both sides to eliminate the square root: ( − 4 x − 6 ) 2 = ( − 2 ) 2 .
Simplifying Simplifying the equation, we get − 4 x − 6 = 4 .
Isolating the x term Adding 6 to both sides, we have − 4 x = 10 .
Solving for x Dividing both sides by -4, we find x = − 4 10 .
Simplifying the Fraction Simplifying the fraction, we get x = − 2 5 .
Checking for Extraneous Solutions Now, we must check for extraneous solutions by substituting x = − 2 5 back into the original equation: − 4 ( − 2 5 ) − 6 = − 2 .
Simplifying the Expression Simplifying the expression under the square root: 10 − 6 = − 2 .
Checking the Solution Further simplifying: 4 = − 2 , which gives 2 = − 2 . Since this is false, there is no real solution.
Final Answer Since the solution x = − 2 5 does not satisfy the original equation, there is no real solution. Therefore, the answer is DNE.
Examples
When dealing with electrical circuits, you might encounter equations involving square roots to determine current or voltage. If solving such an equation leads to a situation where the calculated voltage is negative under a square root, similar to our problem, it indicates that the assumed circuit configuration or parameters are not physically realizable. Checking for extraneous solutions ensures that your calculations align with real-world constraints, preventing incorrect designs or interpretations.
The equation − 4 x − 6 = − 2 has no real solutions because the square root cannot be negative. By squaring and solving, we found x = − 2 5 , but it leads to an invalid equation when substituted back. Therefore, the solution is DNE .
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