Lesson: Quadratic Equations

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@0:33 How did you get x^2 + 7x = 0 ?
gmat-admin's picture

All we're saying here is that x^2 + 7x = 0 is a quadratic equation, since we can rewrite this equation in the form ax^2 + bx + c = 0

In the question 9x^2 + 6x + 1=0, I solved it a different way and still got X=-1/3. I multiplied c*a to get 1x9 =9. Then I expanded 6x to get 9x^2+3x+3x+1=0. I then factored the first half to get 3x(3x+1) and 1(3x+1). I was left with x=-1/3
gmat-admin's picture

That's a valid approach too.

If 2ab - c = 2a(b - c), which of the following must be true?

(A) a=0 and c=0
(B) a=1/2 and b=2
(C) b=1 and c=0
(D) a=1 or b=0
(E) a=1/2 or c=0 The answer is E but I fell it could be c also as I started putting figures with c equation and came equal so I choose c .
gmat-admin's picture

Given: 2ab - c = 2a(b - c)
Expand right side: 2ab - c = 2ab - 2ac
Subtract 2ab from both sides: -c = -2ac
Add 2ac to both sides: 2ac - c = 0
Factor: c(2a - 1) = 0

So, EITHER c = 0, OR 2a - 1 = 0
If 2a - 1 = 0, then a = 1/2

So, either c = 0 or a = 1/2 (answer choice E)

The keyword is MUST. Although b = 1 and c = 0 (answer choice C) is a valid solution, it need not be the case that b = 1. Notice that b = 0 and c = 0 also works.

Thanx for the explaination .

How did you get 169?
gmat-admin's picture

You're referring to 6:05 in the video.

We have 3² - (4)(10)(-4) = 9 - (-160)
= 9 + 160
= 169

Is it possible to load the videos to a laptop or tablet to play/study without an internet connection? If so, how?
gmat-admin's picture

I'm sorry, but our videos are streaming and cannot be downloaded. As such, they're accessible only through an Internet connection.

If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?
Since x = 4 is a solution to the equation x² + 3x + k = 10, we know that x = 4 SATISFIES the equation.

That is: 4² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

Hi Brent,

I did not understand how did you get -18 for K ???

Thanks
Fatima-Zahra
gmat-admin's picture

Let's start here: 16 + 12 + k = 10
Simplify to get: 28 + k = 10
Subtract 28 from both sides of the equation to get: k = -18

Cheers,
Brent

If (x − a)(x − b) = 0, is b = s?

(1) s < a
(2) s is a root of x^2 − ax − bx + ab = 0


Hi Brent, in the above question

(x-a)(x-b) = 0
That means for any value x; a = b
Therefore, the question could be written as "Does a=s?"

1) Using above, this is sufficient
2) Not sufficient because doesn't talk about the value of s

Could you help me identify where have I gone wrong in this?
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-a-x-b-0-is-b-s-228809.html

We cannot conclude that, since (x-a)(x-b) = 0, it must then be the case that a = b

For example, if a = 2 and b = 5, then we have: (x-2)(x-5) = 0
This tells us that EITHER x = 2 OR x = 5
But we can see that a does not equal b.

So, if (x − a)(x − b) = 0, then we can only conclude that EITHER x = a OR x = b.

Cheers,
Brent

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