BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Quadratic equations in DS problems

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 111
Joined: Mon Oct 03, 2011 12:08 pm

Quadratic equations in DS problems

by nonameee » Sat Nov 19, 2011 6:01 am
Quadratic equations can have two positive solutions. In DS questions, this fact can present an obstacle since in order to decide whether the statement is sufficient we need to know whether the quadratic equation yields one or two positive solutions. Now, to get to a quadratic equation you often have to manipulate/simplify algebraic expressions. This alone can be time consuming. To solve a quadratic equation can take you additional time, especially if the equation doesn't have pretty numbers.

I've got several questions:

1) Is there a quick way to know whether a quadratic equation has one or two positive solutions without the need to solve it?

The only thing that comes to mind is to calculate (or rather to approximate) a discriminant and to compare it with b as in: ax^2 + bx + c = 0

2) Is there a quick way to know whether a quadratic equation has whole solutions?

Here we would have to calculate a discriminant, then to compare it with b (EVEN, ODD).

3) Is there a quick way to know whether the following equation has one positive whole solution:

60/n = 60/(n-5) - 2

Thanks.
Top
Source: — Data Sufficiency |
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Sat Nov 19, 2011 11:52 am
i guess the list of your questions may be extended as far as you study quadratics in your GMAT course, but the major learning I have done for myself is to master the concept first, around quadratics. Perhaps you've heard about "FOIL" method which comes as straightforward as you look at the expression. One example is (a+3)^2 OR a^2+6a+9 OR (a+3)(a+3) these all are the expressions of quadratics. Now if you noticed to find the first solution we simply allow for 6a to be written as (3+3) and 9 as 3*3. Whenever you come across quadratics this rule can be applied.

Let's consider your example and decide right away how many solutions we have and are there positive or negative solutions (if you noticed in previous example for one of the solutions to be +ve or -ve we need to have 'mx^2+nx+C=0' be represented as n=(f+g) and C=f*g either one must have +ve or -ve sign ... you can check the signs of your solutions here)

but let's look at your example for the moment given:
60/n = 60/(n-5) - 2 If we rewrite to obtain formal equation with 0 on the RHS we get 60/n-60/(n-5)+2=0 OR 60(n-5) -60n +2(n^2-5n)=0 <--- this is quadratic equation you want to know the solutions right away, correct?

2n^2-10n-60n+60n-300=0 <> 2n^2-10n-300 OR n^2-5n-150=0 (we reduce the a^2 coefficient to 1 for easy "FOIL" rule application)

OK, -5=-15+10 and -150=-15*10 Enough and stop here, no discriminant is needed
rewrite as (a-15)(a+10)=0 and your solution values are a=15, a=-10
you can recheck by applying discriminant too, I haven't done so, because this is good as well

Play around the concepts and analyze the subject and many more intricacies may come to your mind
Success doesn't come overnight!
Top
Master | Next Rank: 500 Posts
Posts: 111
Joined: Mon Oct 03, 2011 12:08 pm

by nonameee » Sat Nov 19, 2011 12:33 pm
Pemdas, thanks for your reply.
OK, -5=-15+10 and -150=-15*10 Enough and stop here, no discriminant is needed
rewrite as (a-15)(a+10)=0 and your solution values are a=15, a=-10
I know this rule. The rule could be applied here because the numbers are "pretty". In fact, to calculate a discriminant here is also easy.

My question concerns more quadratic equations with "ugly" coefficients where the rule you mentioned cannot be applied that easily.

Also, to get a quadratic equation you had to simplify the original expression. Again, with more "difficult" numbers that wouldn't have been as easy (for example, the simplification could have involved tedious multiplications, additions etc.).
Top
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Sat Nov 19, 2011 4:33 pm
as it's said by knowing the signs of 'n' and 'C' of the equation 'mx^2+nx+C=0' you could be able to figure out at least the solution to be represented by +ve or -ve number

don't disgrace me, but I feel you look after philosophical stone here, and there's none

I came to know and find some websites (mostly Indian ones for CAT) which would direct their students to learn various tricks and shortcuts for figuring out the quick ways of solving problems or math situations. You may also research this info, but I am sure what GMAT tests is quite generic approach folded with careful analysis of each concept. 'FOIL' is the only thing I know about quadratics and I also know that we can graphically test the solutions by checking upward/downward direction of graph. But nothing else ...
Success doesn't come overnight!
Top
Master | Next Rank: 500 Posts
Posts: 111
Joined: Mon Oct 03, 2011 12:08 pm

by nonameee » Sun Nov 20, 2011 12:50 am
I came to know and find some websites (mostly Indian ones for CAT) which would direct their students to learn various tricks and shortcuts for figuring out the quick ways of solving problems or math situations. You may also research this info, but I am sure what GMAT tests is quite generic approach folded with careful analysis of each concept. 'FOIL' is the only thing I know about quadratics and I also know that we can graphically test the solutions by checking upward/downward direction of graph. But nothing else ...
I absolutely agree with you on that. But sometimes shortcuts can save you a lot of time. Anyway, thanks for your input.
Top
Post new topic Post Reply
• Page 1 of 1