Lesson: Quadratic Equations

Comment on Quadratic Equations

@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

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
gmat-admin's picture

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/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?
gmat-admin's picture

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-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.
gmat-admin's picture

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

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?
gmat-admin's picture

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?

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
gmat-admin's picture

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"
Answer: C

Does that help?

Cheers,
Brent

what is the right answer for this question:

Is y = 7?
(1) (x - 3) = 0
(2) (x - 3)(y - 7) = 0
gmat-admin's picture

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

Answer: E

Cheers,
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!
gmat-admin's picture

Link: https://gmatclub.com/forum/which-of-the-following-are-roots-of-an-equati...
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,

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
gmat-admin's picture

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

Answer: D

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
gmat-admin's picture

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

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.
gmat-admin's picture

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
We'll return to this later. First, let's learn about Viète's Theorem....
---------------------------
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

Answer: D

Cheers,
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
gmat-admin's picture

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

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

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