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## Comment on

Quadratic Equations## @0:33 How did you get x^2

## All we're saying here is that

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

## That's a valid approach too.

That's a valid approach too.

## If 2ab - c = 2a(b - c), which

(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 .

## Given: 2ab - c = 2a(b - c)

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?

## You're referring to 6:05 in

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

## I'm sorry, but our videos are

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

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

## Let's start here: 16 + 12 + k

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 =

(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?

## Question link: https:/

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

## Hi Brent,

I am stuck on the same question. I don't understand why x=a or x=b.

I thought that x would be the positive value of a or b since they have to be negative.

Could you explain why that is wrong?

Thank you!

Pia

## Question link: https:/

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

Hi Pia,

Even though the equation is in the form (x − a)(x − b) = 0, we don't know whether a or b are negative or positive.

To see why, let's test some values of a and b, and also examine the resulting solutions.

CASE I) a = 3 and b = 5

The equation becomes: (x − 3)(x − 5) = 0

So, EITHER x - 3 = 0, which means x = 3 OR x - 5 = 0, which means x = 5

So, when a = 3 and b = 5, the SOLUTIONS are x = 3 and x = 5

CASE II) a = 3 and b = -1

The equation becomes: (x − 3)(x − (-1)) = 0

Simplify to get: (x − 3)(x + 1) = 0

So, EITHER x - 3 = 0, which means x = 3 OR x + 1 = 0, which means x = -1

So, when a = 3 and b = -1, the SOLUTIONS are x = 3 and x = -1

CASE III) a = -8 and b = -1

The equation becomes: (x − (-8))(x − (-1)) = 0

Simplify to get: (x + 8)(x + 1) = 0

So, EITHER x + 8 = 0, which means x = -8 OR x + 1 = 0, which means x = -1

So, when a = -8 and b = -1, the SOLUTIONS are x = -8 and x = -1

In all cases, the solutions are x = a and x = b

Does that help?

Cheers,

Brent

## Yes, very much!

Thank you!

## https://gmatclub.com/forum

From statement 1 x = 2,-3 (insuff)

Statement 2 is insuff

Combining 1 and 2, if we use x = -3 then there will be no solution because b²- 4ac will be negative.

So x has to be 2

Is this approach correct?

## Question link: https:/

Question link: https://gmatclub.com/forum/what-is-the-value-of-x-1-x-2-x-10-16-2-x-4y-4...

It's hard to tell whether your approach is correct.

When you combine statements 1 and 2, what kind of equation are you getting?

In other words, what equation are you referring to when you mention the discriminant b² - 4ac?

Cheers,

Brent

## https://gmatclub.com/forum/is

Hi Brent could you please clarify the correct answer for this question? https://gmatclub.com/forum/is-y-214659.html

Bunuel has "E" and you state that it is "C".

Appreciate it.

## Thanks for the heads up.

Thanks for the heads up. Looks like I wrote the wrong equation for statement 1 (I wrote x - 4 = 0, when it was actually x - 3 = 0)

I've edited my response accordingly (at https://gmatclub.com/forum/is-y-214659.html#p2057151). Thanks again!

Cheers,

Brent

## Hi Brent,

Need you help:

Let a, b, c, and d be nonzero real numbers. If the quadratic equation ax (cx + d) = –b (cx + d) is solved for x, which of the following is a possible ratio of the 2 solutions?

1 -ab/cd

2. -ac/bd

3. -ad/bc

4. ab/cd

5. ad/bc

## Here's my solution: https:/

Here's my solution: https://gmatclub.com/forum/let-a-b-c-and-d-be-nonzero-real-numbers-if-th...

Cheers,

Brent

## https://gmatclub.com/forum/if

I am having trouble following the reasoning here. What's the simpler way of approaching this problem?

## It's a crazy hard question.

It's a crazy hard question. Here's my solution: https://gmatclub.com/forum/if-x-u-2-v-2-y-2uv-and-z-u-2-v-2-and-if-x-163...

Cheers, Brent

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