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Re: x < 0 [#permalink]
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AE wrote:
Is there any other approach?


You could try testing numbers.

More here:
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Re: x < 0 [#permalink]
AE wrote:
Is there any other approach?



An easier way would be to notice that x^2 will always give a positive value. So on both sides that part is equal.

Moving on to -5x vs -9x for whatever negative value of x the -9x will be larger because (-ve) * (-ve) yields positive.

For example let x be -1 it will be (-1)*(-5) vs (-1)*(-9)

Finally 20 is larger than 6 so on that basis we could tell that B is larger.
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Re: x < 0 [#permalink]
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AE wrote:
Is there any other approach?

Answer: B
x<0
We want to compare x^2 - 5x +6 and x^2 -9x +20, as x^2 is the same amount in both of them, we can omit it and compare - 5x +6 and -9x +20

- 5x +6 and -9x +20
We subtract 6 from both sides, then we will have:
-5x and -9x + 14
As x is something negative, if it’s multiplied by a negative number (-5 in A and -9 in B) it will be positive, So -9x is bigger than -5x. Assume -9x as 9*(-x) and as x is negative, (-x) will be positive, same for -5x)
So B is bigger than A.
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Re: x < 0 [#permalink]
Think plugging in with x= -1 and x=-20, the extreme case will be better
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Re: x < 0 [#permalink]
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AE wrote:
Is there any other approach?


I would approach it this way,

We are to compare A and B. We can subtract B from A and if we can determine whether the resulting expression is -ve, +ve or 0 then we can easily answer which one is greater.
So,
A-B = x^2 - 5x + 6 - x^2 + 9x - 20 = 4x - 20
We are given that x<0 which makes the term 4x always negative. Thus the result above is always negative.
Hence we can come to the conclusion that Quantity A was greater than the Quantity B.
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Re: x < 0 [#permalink]
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Let's try some different approach.

As we already know

if A-B>0 --> A>B
if A-B<0 --> A<B

subtract equation B from Equation A
--> Equation A - Equation B
--> The reduced expression would be
--> 4x-14
--> Since x is negative the expression is negative

So A-B<0 --> B is greater. Hence, B is the Answer.

Please let me know in case of any queries or further breakdown.

Regards,
Sanpreet
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Re: x < 0 [#permalink]
We can try substitution,

x = -0.1

we get LHS = (-2.1)(-3.1) and RHS = (-4.1)(-5.1) . Thus B

if we take x = -2

we get LHS = 20 and RHS 42

Thus B
This method is lengthy however
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Re: x < 0 [#permalink]
1
One way to solve is to factor and then compare based on number
properties. Quantity A factors to (x – 2)(x – 3). Quantity B factors to (x – 4)(x
– 5). Because x is negative, “x minus a positive number” is also negative.
Each quantity is the product of two negative numbers, which is positive:
Quantity A: (x – 2)(x – 3) = (neg)(neg) = pos
Quantity B: (x – 4)(x – 5) = (more neg)(more neg) = more pos
Quantity B is greater.
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Re: x < 0 [#permalink]
1
AE wrote:
Is there any other approach?

You can try my approach:

x<0

QA:

\(x^2-5x+6=(x-3)(x-2)=0\)
Using the number line we know that:
If x>3, x>0
If x<2, x>0
Therefore, the range of QA is 2<x<3

QB
\(x^2-9x+20=(x-5)(x-4)=0\)
Using the number line we know that:
If x>5, x>0
If x<4, x>0
Therefore, the range of QB is 4<x<5

No matter what, there is no overlap value - QB is on the higher range always

B
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Re: x < 0 [#permalink]
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Re: x < 0 [#permalink]
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