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A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
Doing computations we get quantity A equal to \(4x^4-4x^2+3x^3-3x\) and quantity B equal to \(4x^4+4x^2+3x^3+3x\). Now, simplifying we get A equal to \(-4x\) and B equal to \(3\). Then, whatever x between 0 and 1, B is greater!
Doing computations we get quantity A equal to \(4x^4-4x^2+3x^3-3x\) and quantity B equal to \(4x^4+4x^2+3x^3+3x\). Now, simplifying we get A equal to \(-4x\) and B equal to \(3\). Then, whatever x between 0 and 1, B is greater!
a short cut: extract one x from (x^3-x) we get x(x^2-1)(4x+3) on the left. extract one x from (4x^2+3x) we get x(x^2+1)(4x+3) on the right. x(4x+3) canceled out. (x^2-1)<(x^2+1)
correct: A (maybe includes miscalculatings!) (x^3 - x)(4x + 3) = 4x^4 + 3x^3 - 4x^2 -3x
(x^2 + 1)(4x^2 + 3x) = 4x^4 + 3x^3 + 4x^2 + 3x
We compare these two and simplify as much as possible. 4x^4 + 3x^3 - 4x^2 -3x vs 4x^4 + 3x^3 + 4x^2 + 3x Omitting 4x^4 + 3x^3: - 4x^2 -3x vs 4x^2 + 3x -(4x^2 + 3x) vs (4x^2 + 3x) a) When 4x^2 + 3x is negative then the value in the left will be positive when multiplied by it’s negative sign. And the right value will be negative and thus less. 4x^2 + 3x < 0 —> 4x^2 < -3x —> x < -3/4 b) When 4x^2 + 3x is positive the value in the right will be bigger. 4x^2 + 3x > 0 —> 4x^2 > -3x —> x > -3/4
BUT it's mentioned that x is between 0 and 1 and it will be case b. so A is bigger than B
x has no specific value. x can be positive or negative or zero. If so, why the answer is not D since (A) -4x^2-3x and B) 4x^2 + 3x have same value if x=0?
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
We can solve this question using matching operations
x has no specific value. x can be positive or negative or zero. If so, why the answer is not D since (A) -4x^2-3x and B) 4x^2 + 3x have same value if x=0?
The OP posted 0 < x < 1 in the subject line, but forgot to add it to the question. I have since added that information to the question.
Instead of expanding equations check the x^3 - x. x > x^3 as it is fraction. For example take 1/2 then 1/2 > 1/8. So Quantity A is always negative. Quantity B has all positive expressions.
We can simply solve this by not opening the brackets and taking x common on QTY B from the expression (4x^2-3) and then cancelling everything out till we are left with QTY A: x^2-1 QTY B: x^2+1