If a and b are constants, what is the value of a ?

Target question: What is the value of a?

Statement 1:a < b
Definitely NOT SUFFICIENT

Statement 2: (t − a)( t − b) = t² + t − 12
Factor: t² + t − 12 = (t + 4)(t - 3)
Rewrite in terms of (t - a) and (t - b) to get: t² + t − 12 = (t - -4)(t - 3)
There are two possible cases:
Case a: a = -4 and b = 3, in which case a = -4
Case b: a = 3 and b = -4, in which case a = 3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that EITHER a = -4 and b = 3 OR a = 3 and b = -4
Statement 2 tells us that a < b, which means it MUST be the case that a = -4 and b = 3
So, a = -4
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Solution:

We need to determine the value of a.

Statement One Alone:

Knowing a < b does not allow us to determine the value of a. Statement one alone is not sufficient.

Statement Two Alone:

Expanding the left side of the equation, we have:

t^2 - (a + b)t + ab = t^2 + t - 12

Equating like terms from both sides, we have:

-(a + b) = 1 and ab = -12

a + b = -1 and ab = -12

The two numbers that add up to -1 and multiply to -12 are -4 and 3. However, it could be the case that a = -4 and b = 3 or the case that a = 3 and b = -4. Statement two alone is not sufficient.

Statements One and Two Together:

With the two statements, we see that a must be -4 and b must be 3 since a < b. Both statements together are sufficient.

Answer: C
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Ans C:
St. 2: Solving Quadratic, we get
(t-a)(t-b) = (t+4)(t-3)
St. 1: a<b
Therefore, a=-4 and b= +3

GMATPrepNow Could you help me with a small query?

For statement 2



Could we have done this as below and would it be fruitful?

(t − a)( t − b) = t² + t − 12
(t - a) = t² + t − 12
t - a = t² + t − 12
12 = t² + a

and ( t − b) = t² + t − 12
t - b = t² + t − 12
12 = t² + b Therefore since, t² + b = t² + a we get a = b ? (Is this right?)

Thank you
Dablu

I think you're confusing (t − a)( t − b) = t² + t − 12 with (t − a)( t − b) = 0
If (t − a)( t − b) = 0, then we know that either (t − a) = 0 or ( t − b) = 0
We're using the principle that says: If AB = 0, then either A = 0, or B = 0

The same strategy doesn't apply when the quadratic expression does not equal 0.
For example, if AB = 6, we can't conclude that A = 6 or B = 6

So, if (t − a)( t − b) = t² + t − 12, we can't then conclude that either (t - a) = t² + t − 12 or (t - b) = t² + t − 12

Cheers,
Brent

Hey Brent

Since it says in statement B "for all values of t", I just plugged in 0 to get a*b=12 and even combining that with stmt. 1, I figured I can´t get a unique value. Where did I go wrong?

Thanks,

Philipp

Yes, why can we not just plug in 0 for t ?

philippi Moritz279

I will just try to answer what you're asking.

(2) (t−a)(t−b)=t^2+t−12 for all values of t.
If you say t=0
then equation becomes (0-a)(0-b)=0+0-12 which is (-a)(-b)=-12 which is ab=-12. We know that a and b are constants, so values must be 3/4 with one of the two being negative.

TheNightKing

Yes but another solutions could be -2 and 6 for ab=-12.

Moritz279

No. \(t^2+t-12\) still has to hold true. +6-2 gives you 4. You need 1 which is only possible by +4-3.

Asked: If a and b are constants, what is the value of a?

(1) \(a < b\)
Since a and b are unknown
NOT SUFFICIENT

(2) \((t − a)( t − b) = t ^2 + t − 12\), for all values of t.
t=-4 or t=3
a=-4 or 3
NOT SUFFICIENT

(1)+(2)
(1) \(a < b\)
(2) \((t − a)( t − b) = t ^2 + t − 12\), for all values of t.
a=-4 and b=3 since a<b
SUFFICIENT

IMO C

Posted from my mobile device

(1) \(a < b\)

clearly insuff.

(2) \((t − a)( t − b) = t ^2 + t − 12\)

factorise \( t ^2 + t − 12 \) = \((t-4) (t+3)\) ---- roots are (t-4) and (t+3), \(t=+4, -3\)

\((t − a)( t − b) = (t-4) (t+3) \) --- don't be tempted to write a is 4 just because it looks that way. the product could be in any way, so "a" could be 4, or -3


hence, using c we get a<b. thus, a is -3

For this step:

>Rewrite in terms of (t - a) and (t - b) to get: t² + t − 12 = (t - -4)(t - 3)

What is the methodology for getting the -4 and 3?

I understand that a + b = 1 and that a*b = -12

But what is the fastest way to arrive at -4 and 3 if you can't see it immediately?

Hi Bunuel BrentGMATPrepNow ScottTargetTestPrep

I am struggling to understand the yellow highlight when reviewing S2.

If ALL values of t are acceptable, then t = 0 SHOULD WORK as well.
-- If i input t = 0, I get ab = - 12

Thus below image ARE ALL the combinations available for a.b = -12

Thus I chose E because if I combine the statements, the green highlights in my list work for the value of A when statements are combined as well.

You're correct to conclude that ab = -12, which means there are infinitely many pairs of values that satisfy this equation (a = 2 & b = -6, a = 0.5 & b = -24, etc)

Since (t − a)( t − b) = t² + t − 12, we can expand and simplify the left side to get: t² - at - bt + ab = t² + t − 12

Factor part of the left side to get: t² + t(-a - b)+ ab = t² + 1t − 12

At this point, we can see that ab = −12 AND (-a - b) = 1

We can take the equation -a - b = 1 and multiply both sides by -1 to get: a + b = -1

Now that we have two pieces of information, ab = −12 AND a + b = -1, we can conclude that a = -4 and b = 3

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It's also worth noting that inputting t = 0 helped you arrive at the conclusion that ab = -12
If you plug in another value for t, you'll get additional information about the values of a and b.
For example, if t = 1, we get: (1 − a)( 1 − b) = 1² + 1 − 12
Simplify: 1 - a - b + ab = -10
Subtract 1 from both sides: a - b + ab = -11. This equation further restricts the values of a and b.

And so on......

Does that help?

Just as you said, the given equality should hold for all values of t. So, for instance, it should hold for t = 1. If you substitute t = 1 into the given equation, you'll obtain (1 - a)(1 - b) = -10, which is equivalent to ab - a - b = -11. If you substitute a = -12 and b = 1, for instance, you'll see that ab - a - b = -11 is not satisfied. That eliminates one column. Using this process (with more values of t, if necessary), you will be able to eliminate every green highlighted column in your table besides a = -4, b = 3.
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