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

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

-------------------------

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.

­Frankly, I still dont get the point about t being equal to 0. Isnt t a varialbe that can be any defined set of numbers? So t could be all real numbers or 0. 

Since we dont know how t is deifned, we cannot rule out the case that t=0 and only 0. 

I guess the dicussion here is about the interpretation of "for all values of t." To me, this means all values that t is defined for, and there is no information on what those values are. To arrive to the correct answer, I guess you must assume that t can take any value, which I dont think is clear based on the statement in the question. 

 

­
The statement "\((t − a)( t − b) = t ^2 + t − 12\), for all values of t" implies that this equation must hold true regardless of the value of t. How is this possible? When we factor the right-hand side of the equation, we get: \((t − a)( t − b) = (t - 3)(t + 4)\). Notice that for this equation to be hold for all values of t, a must be 3 and b must be -4, or vice versa. In either case, we end up with \((t − 3)( t + 4) = (t - 3)(t + 4)\), which holds true for all values of t: -17.4, 0, 1, 100, and so forth. Therefore, from statement (2), a can be either 3 or -4. When combining the statements, given that a < b, we deduce that a = -4 and b = 3. Therefore the answer is C.

Hope it helps.

­Okay, that makes complete sense. Thank you!­