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If a#b = a^2-ab, which of the following expressions cannot
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21 May 2020, 10:41
21 May 2020, 10:41
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If \(a#b = a^2 - ab\), which of the following expressions cannot be negative?
Indicate all such expressions.
❑ x#(x - 1)
❑ x#0.5x
❑ |x|#|y|
❑ |x-y|#-2
❑ 0#(x + 1)
❑ (-x- 2)#0
Indicate all such expressions.
❑ x#(x - 1)
❑ x#0.5x
❑ |x|#|y|
❑ |x-y|#-2
❑ 0#(x + 1)
❑ (-x- 2)#0
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Official Answer
B,D,E,F
Re: If a#b = a^2-ab, which of the following expressions cannot
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21 May 2020, 11:20
21 May 2020, 11:20
2
❑ x#(x - 1) = x*x - x*(x-1) = x (can be negative if x < 0)
❑ x#0.5x = x*x - 0.5*x*x = 0.5*x*x (always non-negative)
❑ |x|#|y| = |x|*|x| - |x|*|y| = |x|*(|x| - |y|) (can be negative if |y| > |x|)
❑ |x-y|#-2 = |x-y|*|x-y| - (-2)*|x-y| = |x-y|*(|x-y|+ 2) (always non-negative)
❑ 0#(x + 1) = 0 (obvious)
❑ (-x- 2)#0 = (-x- 2)*(-x- 2) - 0 (always non-negative)
❑ x#0.5x = x*x - 0.5*x*x = 0.5*x*x (always non-negative)
❑ |x|#|y| = |x|*|x| - |x|*|y| = |x|*(|x| - |y|) (can be negative if |y| > |x|)
❑ |x-y|#-2 = |x-y|*|x-y| - (-2)*|x-y| = |x-y|*(|x-y|+ 2) (always non-negative)
❑ 0#(x + 1) = 0 (obvious)
❑ (-x- 2)#0 = (-x- 2)*(-x- 2) - 0 (always non-negative)
Re: If a#b = a^2-ab, which of the following expressions cannot
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25 May 2020, 15:15
25 May 2020, 15:15
1
Another way to look at this is to see that as long as a^2 is greater than a*b, the expression cannot be negative. This should save you time because you won't have to solve every single line and can stick to either using number sense or plugging numbers depending on what's easier.
Let's have a look -- all of the below correlates to: a^2 & a*b
(a) x^2 & x^2 - x ----> if x is positive, then a^2 might be more, but if x is negative, then a^2 might be less. NOT AN ANSWER.
(b) x^2 & 0.5x^2 ----> half of a square is always going to be less than the square. THIS IS AN ANSWER.
(c) x^2 & xy -----> Like, come on, we don't know what the heck y even is. It could be anything and could make xy lesser or greater than x^2. NOT AN ANSWER.
(d) (x-y)^2 & (-2)*|x-y| ---> first off, that a^2 is going to be the value of x-y squared makes the absolute value of it moot. However, in the second expression, the absolute value function makes sure that the second expression is, indeed, negative (as per multiplying by -2). That means that the second expression will always be less than the first. THIS IS AN ANSWER
(e) 0 & 0*(x+1) ---> 0 and 0. This will be non-negative. THIS IS AN ANSWER
(f) (-x-2)^2 & 0 ----> 0 will always be less than (-x-2)^2, as the first expression is squaring a value thereby making sure that that first expression is never negative. THIS IS AN ANSWER
Always look for ways to apply logic to do less work where possible!
Let's have a look -- all of the below correlates to: a^2 & a*b
(a) x^2 & x^2 - x ----> if x is positive, then a^2 might be more, but if x is negative, then a^2 might be less. NOT AN ANSWER.
(b) x^2 & 0.5x^2 ----> half of a square is always going to be less than the square. THIS IS AN ANSWER.
(c) x^2 & xy -----> Like, come on, we don't know what the heck y even is. It could be anything and could make xy lesser or greater than x^2. NOT AN ANSWER.
(d) (x-y)^2 & (-2)*|x-y| ---> first off, that a^2 is going to be the value of x-y squared makes the absolute value of it moot. However, in the second expression, the absolute value function makes sure that the second expression is, indeed, negative (as per multiplying by -2). That means that the second expression will always be less than the first. THIS IS AN ANSWER
(e) 0 & 0*(x+1) ---> 0 and 0. This will be non-negative. THIS IS AN ANSWER
(f) (-x-2)^2 & 0 ----> 0 will always be less than (-x-2)^2, as the first expression is squaring a value thereby making sure that that first expression is never negative. THIS IS AN ANSWER
Always look for ways to apply logic to do less work where possible!
