@0:33 How did you get x^2

@0:33 How did you get x^2 + 7x = 0 ?

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

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

That's a valid approach too.

That's a valid approach too.

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

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 .

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 .

Thanx for the explaination .

How did you get 169?

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

Is it possible to load the videos to a laptop or tablet to play/study without an internet connection? If so, how?

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

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

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 =

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?

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,

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

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

https://gmatclub.com/forum/what-is-the-value-of-x-1-x-2-x-10-16-2-x-4y-4-2y-212386.html

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?

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

https://gmatclub.com/forum/is-y-214659.html

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. 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,

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

4. ab/cd

https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-x-u-2-v-2-y-2uv-and-z-u-2-v-2-and-if-x-163805.html

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

For Is x>0?

For Is x>0?

1) x^2=9x
2) x^2=81

Can i solve x^2=9x as x*x= 9x
So x= 9.
Ans will be A

Many students will fall for

Many students will fall for the same trap that you fell for.

Once we know that (x)(x) = (9)(x), it's very tempting to divide both sides by x to get: x = 9
HOWEVER, this reasoning is incorrect, since we don't know whether x = 0
If x = 0, then we are dividing by 0, which causes many problems.

In fact, we can see that the equation x² = 9x has TWO solutions: x = 9 and x = 0

To avoid the possibility of accidentally dividing by zero, we can treat the equation, x² = 9x, as a quadratic equation.
Take: x² = 9x
Subtract 9x from both sides to get: x² - 9x = 0
Factor to get: x(x - 9) = 0
So, EITHER x = 0 OR x = 9
If x = 0, then x is NOT greater than 0
If x = 9, then x IS greater than 0
So, statement 1 is not sufficient.

Statement 2: x² = 81
Solve to see that EITHER x = 9 OR x = -9
If x = 9, then x IS greater than 0
If x = -9, then x is NOT greater than 0
So, statement 2 is not sufficient.

When we combine the two statements, we can conclude that x MUST equal 9
So, the answer to the target question is "YES, x IS greater than 0"

Does that help?

Cheers,
Brent

what is the right answer for

what is the right answer for this question:

Is y = 7?
(1) (x - 3) = 0
(2) (x - 3)(y - 7) = 0

TARGET QUESTION: Is y = 7?

TARGET QUESTION: Is y = 7?

STATEMENT 1: (x - 3) = 0
There's no information about y, so there's no way to determine whether or not y = 7
Since we cannot answer the TARGET QUESTION with certainty, statement 1 is NOT SUFFICIENT
--------------------
STATEMENT 2: (x - 3)(y - 7) = 0
This tells us that EITHER x = 3 OR y = 7

Consider these two cases:

CASE A: x = 2 and y = 7. This is a solution to the given equation. In this case, the answer to the target question is "YES, y DOES equal 7)

CASE B: x = 3 and y = 1. This is a solution to the given equation. In this case, the answer to the target question is "NO, y does NOT equal 7)

Since we cannot answer the TARGET QUESTION with certainty, statement 2 is NOT SUFFICIENT
--------------------
STATEMENTS COMBINED
There are still many possible values of x and y that satisfy BOTH statements.
Consider these two cases:

CASE A: x = 3 and y = 7. This is a solution to the given equation. In this case, the answer to the target question is "YES, y DOES equal 7)

CASE B: x = 3 and y = 1. This is a solution to the given equation. In this case, the answer to the target question is "NO, y does NOT equal 7)

Since we cannot answer the TARGET QUESTION with certainty, the COMBINED statements are NOT SUFFICIENT

Cheers,
Brent

Hi Brent,

Hi Brent,

There are equations in this section such as 15x^2-2x-1, 2x^2+9x+9, where a=15, a=2, which are not perfect squares, hence I cannot really apply formula (a+b)^2=a^2+2ab+c^2, and neither they fit (x+a)(x+b) = x^2+nx+p (because in order to apply there should not be anything in front of x^2). So I am puzzled here how to quickly factor such type of equations. In your solution, you give 15x^2-2x-1 =(5x+1)(3x-1) https://gmatclub.com/forum/which-of-the-following-are-roots-of-an-equation-x-2-2x-224086.html Do you have a video explanation how to factor this type of equations other then use formula (-b+/-square root of (b^2-4ac))/2a??
Thanks a million!

Sorry for the delay. I don't know how this questions eluded me.

Great question.
IF, on the GMAT, you get a quadratic expression that's not in the form x² + ax + b = 0 or in the form of a special product (e.g., x² + 2xy + y²), then then coefficients in the expression will be such that you'll be able to factor by testing a very small number of cases.

Take, for example, the expression 15x² - 2x - 1
We know that the two constants must multiply to get -1
So, the constants must be 1 and -1
So, we have: 15x² - 2x - 1 = (? + 1)(? - 1)

We also know that the first expressions (denoted by ?'s) must multiply to get 15x²
So, those first expressions are EITHER 15x & x OR 5x & 3x

Let's test the first pair
Is it true that 15x² - 2x - 1 = (15x + 1)(x - 1)?
Upon expanding the right side, we see that it is NOT the case that 15x² - 2x - 1 = (15x + 1)(x - 1)

Now test the second pair
Is it true that 15x² - 2x - 1 = (5x + 1)(3x - 1)?
Upon expanding the right side, we see that it IS the case that 15x² - 2x - 1 = (5x + 1)(3x - 1)

DONE!

Cheers,
Brent

Hey Brent,

Hey Brent,

how would you solve this questions with the methods we have learned?

I honestly had quite a hard time following up on the answer explanations on the forum.

https://gmatclub.com/forum/which-of-the-following-equations-has-1-2-as-one-of-its-roots-220316.html

Cheers,

Philippon

https://gmatclub.com/forum

https://gmatclub.com/forum/which-of-the-following-equations-has-1-2-as-o...

This is a very tricky question!!!
One option is to apply the quadratic formula to all 5 answer choices and see which one yields a solution of x = 1 + √2

Alternatively, if we examine the solutions to the quadratic formula, we get solutions in the form x = k +/- √(some number)

That is, if x = k + √(some number) is one solution, then the other will be x = k - √(some number)

So, if x = 1 + √2 is one solution, then x = 1 - √2 is another solution.

-------------------------------------
ASIDE: Before we continue, notice that, if x = 3 and x = 2 are two solutions to an equation, then we know that the equation looks something like (x - 3)(x - 2) = 0 (notice that x = 3 and x = 2 are the two solutions )

Likewise, if x = -4 and x = 9 are two solutions to an equation, then we know that the equation looks something like (x + 4)(x - 9) = 0 (notice that x = -4 and x = 9 are the two solutions )
-------------------------------------

So, if x = 1 + √2 and x = 1 - √2 are two solutions to an equation, then we know that the equation looks something like:

[x - (1 + √2)][x - (1 - √2)] = 0
Simplify: (x - (1 - √2)(x - 1 + √2) = 0
Expand and simplify: x² - 2x - 1 = 0

Does that help?

Cheers,
Brent

Thanks for the answer Brent! This one is very similar I guess, how would yopu tackle it?

Does it make sense to be familiar with the mentioned Theorem?

Cheers

Hi Philippi,

Hi Philippi,
Did you mean to include a question link?

I am sorry...I indeed meant

I am sorry...I indeed meant to include a link, now I don´t remember which one it was. In any case, I figured that the Viete Theorem is basically what we were doing. Having the two roots, the Viete Theorem tells us that the equation must be x^2-bx+c, correct?

Cheers.

I'm not a big fan of having

I'm not a big fan of having students memorize Viète's Theorem, BUT let's check it out.

When we apply (above) what we know about the quadratic formula, we concluded that the 2 solutions to the unknown equation will be x = 1 + √2 and x = 1 - √2
---------------------------
When applied to quadratic equations, Viète's Theorem says:
If x1 and x2 are solutions to the equation ax² + bx + c = 0, then:
x1 + x2 = -b/a
and
(x1)(x2) = c/a

IMPORTANT: When we examine the 5 answer choices (at https://gmatclub.com/forum/which-of-the-following-equations-has-1-2-as-o...) we see that the a-value for each quadratic equation is 1.

Since we're basically told that a = 1, we can now take Viète's Theorem and replace a with 1 to get:
If x1 and x2 are solutions to the equation x² + bx + c = 0, then:
x1 + x2 = -b
and
(x1)(x2) = c
------------------------------

Now we'll return to the fact that we know the 2 solutions to the mystery equation will be x = 1 + √2 and x = 1 - √2

Let x1 = 1 + √2 and let x2 = 1 - √2

Applying the part of Viète's Theorem that says x1 + x2 = -b, we can write: (1 + √2) + (1 - √2) = -b
Simplify to get: 2 = -b
So, b = -2

Applying the part of Viète's Theorem that says (x1)(x2) = c, we can write: (1 + √2)(1 - √2) = c
Simplify to get: 1 - 2 = c
Simplify: -1 = c

So, if b = -2 and c = -1, our equation x² + bx + c = 0 becomes x² - 2x - 1 = 0

Cheers,
Brent

Hi Brent,

Hi Brent,

is this a realistic GMAT question?

https://gmatclub.com/forum/if-x-2-2x-15-x-r-x-s-for-all-values-of-x-and-if-r-and-210263.html

I think it's realistic. I'd say it's in the 650-700 range.

Hi Brent,

Hi Brent,

i noticed that when I use the factorization method for the equation 10x^2 + 3x - 4 = 0, i got (x-5)(x+8). so why is it different from the answer you got from the formula?

If we expand and simplify (x

If we expand and simplify (x-5)(x+8), we get x² + 3x - 40. So, we know that's not right.

The factorization technique you're using applies only to quadratic equations in the form x² + jx + k = 0 (the coefficient of the x² term is 1)
In 10x² + 3x - 4 = 0, the coefficient of the x² term is 10. So, we can't apply the same technique.

To factor 10x² + 3x - 4 = 0, I suggest that you test a few values (the GMAT won't give you a crazy quadratic to factor, so you won't have to test many options).

First consider what two expressions have a product of 10x².
There are only two pairs expressions that give us this product: 2x & 5x AND 10x & x
Let's first test 2x & 5x
We have (2x )(5x )

Now let's consider what two numbers have a product of -4
What two numbers COULD be 2 and -2. So, one possible option is (2x + 2)(5x - 2)
When we expand and simplify this we get: (2x + 2)(5x - 2) = 10x² + 6x - 4 NO GOOD.

I'll try reversing the placement of the 2 and -2 to get: (2x - 2)(5x + 2)
When we expand and simplify this we get: (2x - 2)(5x + 2) = 10x² - 6x - 4 NO GOOD.

But that's another two numbers have a product of -4
So, one possible option is (2x + 1)(5x - 4)
When we expand and simplify this we get: (2x + 1)(5x - 4) = 10x² - 3x - 4 CLOSE!
Since we have -3x instead of +3x, this probably means that we need to just change the signs of our numbers.

Let's try (2x - 1)(5x + 4)
When we expand and simplify this we get: (2x - 1)(5x + 4) = 10x² + 3x - 4 PERFECT!!!

NOTE: On test day, it is highly unlikely that you will be asked to factor a quadratic in which the coefficient of x² is not 1.
However, in such a case, you want to meet to test many values in order to factor the expression.

thank you for the explanation

thank you for the explanation.

Hi Brent,

Hi Brent,

For below question:
If (1−p) is a root of quadratic equation x^2 + px + (1−p) = 0 then its roots are

A. 0, -1
B. -1, 1
C. 0, 1
D. -1, 2
E. 2, 3

1-p is a factor of the equation.
Can I take it as 1-p = 0 and p = 1 to get the value of p and apply in the same equation to get value of X.

Thanks,
Dev

Good question!

Good question!

If 1-p is a root of an equation, we can't automatically conclude that 1-p = 0 (although this happens to be the case with this question)

Consider this rudimentary question:
If (1−p) is a root of equation x - 5 = 0 then its root is....?

If (1−p) is a solution (root) we can replace x with 1-p to get: (1-p) - 5 = 0
This tells us that 1-p = 5
So, as you can see, knowing that 1-p is a root of the equation does not necessarily mean that 1-p = 0

Cheers, Brent

Hi Brent,

Hi Brent,

Can you please comment on the below data sufficiency question:

Is z>2?
1) |(z-2)^2|>4
2) |8x-4k|=z

For part 1, I got:
(z-2)^2>4
z^2-4z+4-4>0
z^2-4z>0
z(z-4)>0
Therefore z>0 or z>4

However, the solution shows z<0 or z>4. Why is that?

Thanks

You're correct to say that: z

You're correct to say that: z(z-4) > 0
This means the critical points are z = 0 and z = 4

Let's examine the following three regions:

Region i: z < 0
If z < 0, then z is negative and (z-4) is negative, which means z(z - 4) > 0. Perfect!
So, values of z in the region z < 0 ARE solutions did the given inequality

Region ii: 0 < z < 4
If 0 < z < 4, then z is positive and (z-4) is negative, which means z(z - 4) < 0. No good.
So, values of z in the region 0 < z < 4 are NOT solutions to the given inequality.

Region iii: 4 < z
If 4 < z, then z is positive and (z-4) is positive, which means z(z - 4) > 0. Perfect!
So, values of z in the region 4 < z ARE solutions to the given inequality.

For more on solving quadratic inequalities, you can watch the following video: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

Hi Brent,

Hi Brent,

I understand your reasoning here: Region i: z < 0
If z < 0, then z is negative and (z-4) is negative, which means z(z - 4) > 0. But why are we considering z<0 and not z>0?Is it because we need to look at it as points on a line and then we have two points 0 and 4? Anything below o, anything in between 0 and 4 and anything above 4? For all quadratic equations then, do we have to look at it in this way? We can't just take z(z-4)>0 as z>0 and Z-4>0?

We know that z = 0 and z = 4

We know that z = 0 and z = 4 are solutions to the EQUATION z(z-4) = 0

These two values (0 and 4) divide the number line into three separate regions:
z < 0, 0 < z < 4, and 4 < z

This means we must examine what happens when z is in each of those three regions.
First, we examined what happens when z is in the first region (z < 0)

Then, we examined the other two regions.
So, to answer your question, we are indeed examining what happens when z > 0, but we're just examining two different versions: 0 < z < 4, and 4 < z

We need to examine these two ranges separately, because z(z - 4) doesn't behave the same way when 0 < z < 4 as it does when z > 4

If you haven't yet had a chance to review the video on quadratic inequalities, take a look: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

You'll find that everything is explained there.

Thanks Brent, this helps. So

Thanks Brent, this helps. So for any quadratic inequality, the first thing we need to do is set the equation equal to zero, right?