Solve the following algebraic equation: #|2x^2-x|=2x^2-x+8#?

Answer 1

#x =1/4 (1 pm i sqrt[31]) #

Making #y = 2x^2-x# we have
#abs y - y = 8#
with solution #y = -4#

Now, in keeping with

#y = 2x^2-x = -4#

We possess the answers.

#x =1/4 (1 pm i sqrt[31]) #
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Answer 2

To solve the equation |2x^2-x|=2x^2-x+8, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative.

Case 1: 2x^2-x is positive: 2x^2-x = 2x^2-x+8 This simplifies to 0 = 8, which is not true. Therefore, there are no solutions in this case.

Case 2: 2x^2-x is negative: When 2x^2-x is negative, we need to change the sign of the expression inside the absolute value. -(2x^2-x) = 2x^2-x+8 This simplifies to -2x^2+x = 2x^2-x+8 Rearranging the terms, we get 4x^2 = 9 Taking the square root of both sides, we have x = ±√(9/4) Therefore, the solutions to the equation are x = -3/2 and x = 3/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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