First of the quadratic equation is incorrect.did you mean X^2 -5X - 6 = 0
Anyway , always remember these 2 formulas
(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2
this is partially help

Really, if I can get this one answered, in detail, it will help resolve the entire theory behind it:

x^2-x-2=0
becomes
(x-2)(x+1)=0

Therefore, and I get this part... X could be 2 or -1.

dapplegate wrote:Really, if I can get this one answered, in detail, it will help resolve the entire theory behind it:

x^2-x-2=0
becomes
(x-2)(x+1)=0

Therefore, and I get this part... X could be 2 or -1.

Good formula to know... that helps with the third part.

dapplegate wrote:Really, if I can get this one answered, in detail, it will help resolve the entire theory behind it:

x^2-x-2=0
becomes
(x-2)(x+1)=0

Therefore, and I get this part... X could be 2 or -1.

Good formula to know... that helps with the third part.

The reason the equations you referenced above become what they became is because of a process called factoring the equations. You're in essence asking your self, "what two things multiply to get this equation?" The following website will be you understand this and it is relatively brief.

https://www.purplemath.com/modules/solvquad.htm

This is a very important topic on the GMAT and a useful tool to solve a variety of problems. It will really increase your score if you have a solid understanding of how this works. If you want a more thorough take on the factoring equations I would suggest Manhattan GMAT's Equations, Inequalities, and VICs strategy guides. It will go over this both in an overview and in an advanced basis.

Hope that helps.

Thanks,

Jared

x2-5x-6=0
x2-6x+x-6=0
x(x-6)+(x-6)
(x-6)(x+1)

2nd

n2-14n+49=0
n2-(7+7)n+49
n2-7n-7n+49=0
n(n-7)-7(n-7)
(n-7)(n-7)=0

(x-2)2
is simple rule (a-b)2=a2+b2-2ab

dapplegate wrote:HOW DOES?
X^2-5X-X=0 become (X-6)(X+1)=0

AND

n^2-14n+49=0 become (n-7)(n-7)=0

AND

(X-2)^2 become X^2 -4X +4

these are simple question of factorization and you have to solve some of these problems to actually master these type of problems.

ok going with first question
X^2-5X-6=0 so here first check for for -6(which is constant term and factorize is so we get -6 = -1 x 2 x 3 x1
so we start making combination of factors in two such that their sum = middle term
here we find -6 x 1= -6(constant term) and -6 +1 = -5 (coeff. of x)
so we try to arrange the given terms that way X^2-5X-6=0 --> X^2-6X + X-6=0 --> X(X-6) + 1(X-6) =0 --> (X-6) (X+1) =0

2nd question
n^2-14n+49=0
again the same process
n^2-14n+49=0 --> n^2-7n-7n 49=0 --> (n-7) (n-7)=0

rather u can solve it by a simple formulae
(a-b)^2 = (a-b)(a-b) = (a^2- 2ab - b^2)


3rd question
again use the same formulae for this question and you will get the result
(X-2)^2= X^2 -4X +4

pradeepkaushal9518 wrote:x2-5x-6=0
x2-6x+x-6=0
x(x-6)+(x-6)
(x-6)(x+1)

2nd

n2-14n+49=0
n2-(7+7)n+49
n2-7n-7n+49=0
n(n-7)-7(n-7)
(n-7)(n-7)=0

(x-2)2
is simple rule (a-b)2=a2+b2-2ab

@pradeep bhai,

Plz use ^ to mean power(s). It will be helpful for the user*

You guys rock. I got a 640 on my first MGMAT practice test, but I always solve the "simple" algebra w/substitution. My quant wasn't the best either. I'm "good" at math, but obviously forgot how to do 7th grade basics!

This makes so much sense now...

dapplegate wrote:You guys rock. I got a 640 on my first MGMAT practice test, but I always solve the "simple" algebra w/substitution. My quant wasn't the best either. I'm "good" at math, but obviously forgot how to do 7th grade basics!

This makes so much sense now...

may I doubt?