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## Comment on

Quadratic Equations## What happens if (b^2-4ac) is

## It makes no difference

It makes no difference whether (b² - 4ac) is odd or even.

If (b² - 4ac) is negative, the quadratic equation has no solution. The reason for this is that we cannot find the square root of ANY negative value.

## http://gre.myprepclub.com/forum

In this link, I don't understand how Carcass came to answer "D"?

I got "C".

## I have provided a step-by

I have provided a step-by-solution here: https://gre.myprepclub.com/forum/x-3-y-1898.html#p14590

I hope that helps.

Cheers,

Brent

## So as long as either one the

so to say you can have x = 9999 and y = 4 (9999)(0) = 0. Simple becuase the equation has to equal 0 to be correct?

## Yes, that's correct.

Yes, that's correct.

As long as a pair of values (x = something, and y = some number) satisfies the equation (x + 3)(y - 4), then that pair of values is a solution to the equation.

Cheers,

Brent

## Hello, I have a question

Explanation part I don't understand is that " then simply by noting that the quadratic polynomial x^2-2x+1 can be factored "

I thought if there's "quant a" and "quant b", we had to equally simply the same amount. But the explanation seems like 2x went to the "quant a" side out of no where.

I will kindly wait for your explanation.

Also, please let me know where I can post question regarding ETS question. Is it fine to comment here ?

## This is the perfect place to

This is the perfect place to ask this question.

Given:

Quantity A: x² + 1

Quantity B: 2x - 1

This question requires us to first recognize that we have a nice quadratic expression "hiding" within the two quantities.

If we recognize that x² - 2x + 1 can be factored into (x - 1)(x - 1), then we might try to create the expression x² - 2x + 1

This can be accomplished by subtracting 2x from BOTH quantities to get:

Quantity A: x² - 2x + 1

Quantity B: -1

ASIDE: That last step is part of the matching operations strategy described in this video: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...

At this point, we can use the fact that x² - 2x + 1 = (x - 1)(x - 1) to rewrite quantity A:

Quantity A: (x - 1)(x - 1)

Quantity B: -1

This is the same as writing:

Quantity A: (x - 1)²

Quantity B: -1

Since Quantity A is the SQUARE of some value, we know that Quantity A must be greater than or equal to zero.

So, Quantity A is definitely greater than -1 (aka Quantity B)

Here's my step-by-step solution to the above question: https://gre.myprepclub.com/forum/x-6318.html#p11241

Does that help?

Cheers,

Brent

## Is the correct answer D (19)

Thanks!

## Good catch - I forgot to

Good catch - I forgot to include the correct answer in my post.

Yes, the answer is D.

Here's my full solution: https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the...

Cheers,

Brent

## Thanks for the quick response

## For this one: https:/

It says that x < 0. The quantities A and B are not equations individually. However, I calculated the roots of the two quantities by solving them as equations and compared their roots. I also got the correct answer.

Do you think this is also a correct approach?

Thanks!!

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/x-5964.html

Unfortunately, it was a coincidence that you got the correct answer.

The reason for this is that (I assume) you arbitrarily set each expression equal to 0, when you could have also set the equations equal to 10 or -999.

Consider this rudimentary example:

QUANTITY A: 2x - 6

QUANTITY B: x + 2

If x = 1, then quantity B is greater. If x = 10, then quantity A is greater. If x = 8, then the quantities are equal.

So, the correct answer is D

However, if we create two equations by arbitrarily setting each expression equal to 0, we get:

2x - 6 = 0 and x + 2 = 0

When we solve them we get x = 3 and x = -2, respectively.

So, if we apply your approach, we incorrectly get A as our answer.

Does that help?

Cheers,

Brent

## Yep, makes sense. Thank you!!

## https://gre.myprepclub.com/forum

x^4 - 2x^2 + 9x^2 = 54

I squared the entire equation to get:

x^4 + 7x^2 - 54 = 0

and used a = x^2 and continued with factorization

then I used -b+-√b^2 - 4ac / 2a formula

subsequently I was getting the wrong answer. Is this approach incorrect?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-then-x-11490.html

The problem with your approach is that we can't just square each individual part on each side.

Instead, we must square each side in its ENTIRETY.

I explain this in much greater detail in my post below.

Cheers,

Brent

## https://gre.myprepclub.com/forum

x^(2/3) - x^(1/3) - 2 = 4

I cubed the whole equation to get rid of the power signs to get x^2 - x - 72 = 0.

When I solved this equation I got the answer of 9 and -8 which is the wrong answer.

What's wrong with this approach?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the...

If we're going to CUBE each side of the equation, we can't just cube each individual part on each side.

Here's an illustrative example: 1 + 1 + 1 = 3 (TRUE!)

Now let's CUBE each part to get: 1³ + 1³ + 1³ = 3³

Evaluate to get: 1 + 1 + 1 = 27 (NOT TRUE)

Likewise, if we have the equation x + 1 = 5, we cannot just square each individual part to get: x² + 1² = 5²

If we do that, we'll get a much different solution than we'd get if we just solve the equation x + 1 = 5

So, if we want to cube each side, we must cube each side in its ENTIRETY. Here's what I mean:

Start with: 1 + 1 + 1 = 3 (TRUE!)

Cube both sides: (1 + 1 + 1)³ = 3³

Simplify: (3)³ = 3³ TRUE!

Likewise, if we start with: x^(2/3) - x^(1/3) - 2 = 4

We can cube both sides as follows: [x^(2/3) - x^(1/3) - 2]³ = 4³

In other words: [x^(2/3) - x^(1/3) - 2][x^(2/3) - x^(1/3) - 2][x^(2/3) - x^(1/3) - 2] = 4³

As you can imagine, we do NOT want to try to expand and simplify the mess on the left side.

So, we need to try a different approach.

Does that help?

Cheers,

Brent

## For our first step, let's

For our first step, let's eliminate the fraction by multiplying both sides by 2x to get:

8x^21 + 12x^20 - 108x^19 + √(36x^4) = 6x^2

Now simplify √(36x^4) to get: 8x^21 + 12x^20 - 108x^19 + 6x^2 = 6x^2

Subtract 6x^2 from both sides to get: 8x^11 + 12x^20 - 108x^19 = 0

Factor out 4x^19 to get: 4x^19(2x^2 + 3x - 27) = 0

Factor to get: 4x^19(2x + 9)(x - 3) = 0

So, it's possible that 4x^19 = 0, in which case x = 0*

Or 2x + 9 = 0, in which case x = -4.5

Or x - 3 = 0, in which case x = 3

I've copied one of your solutions to the practice questions linked to this page. The questions was to identify the values of x satisfied the original equation. But, my question is about the square root conducted early in the steps. I'm confused as to why there are not 2 solutions when simplified (a positive and a negative)? I've noticed in other problems that sometimes only one solution is given and sometimes 2 solutions are given. And I thought completing square roots always resulted in 2 solutions (a positive and a negative). What is the reasoning behind only giving one solution in a problem like this one?

## Hi e-steele,

Hi e-steele,

In the future, please include the link, so it's easier for me to find the question.

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-3-which-of-the-follo...

The main idea here is that the square root NOTATION (√) directs us to single out the POSITIVE square root of a value.

So, for example, √4 = 2 (not -2)

Likewise, √81 = 9, and √49 = 7

Conversely, the equation x² = 25 does not have any square root NOTATION.

In this case, there are two solutions: x = 5 and x = -5

Likewise, if y² = 36, then EITHER y = 6 OR y = -6

In GENERAL, we can write:

If x² = k, then EITHER x = -√k OR x = -√k

Does that help?

Cheers,

Brent

## https://gre.myprepclub.com/forum

For this question how did you get the value of x=0 and 2 from x(x-2)=0

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/solve-x-x-2-x-3-x-2154.html

At 1:40 in the above video, we learn that, if AB = 0, then either A = 0 or B = 0

So, if we know that (x)(x - 2) = 0, then it must be the case that either x = 0 or (x - 2) = 0

Does that help?

Cheers,

Brent

## Jeez forgot about that.

## X^4=29x^2-100

How can this be solved to all possible values of x

## We need to recognize that

We need to recognize that this is a quadratic equation in disguise.

NOTE: This is a pretty tricky question, and the following technique is advanced as well.

Let k = x²

This means that k² = (x²)² = x⁴

So, upon replacement, the equation x^4 = 29x^2 - 100 becomes: k² = 29k - 100

Rearrange the equation to get: k² - 29k + 100 = 0

Factor to get: (k - 4)(k - 25) = 0

This means either (k - 4) = 0 or (k - 25) = 0

If k - 4 = 0, then k = 4

If k - 25 = 0, then k = 25

So, we have two solutions: k = 4 and k = 25

But wait! We're not done.

We're asked to find the value of x (not k)

So, now we'll go back and replace k with x².

We get: x² = 4 and x² = 25

If x² = 4, then x = 2 or x = -2

If x² = 25, then x = 5 or x = -5

So, there are four possible solutions: x = 2, x = -2, x = 5 and x = -5

Cheers,

Brent

## https://gre.myprepclub.com/forum

How is 4x^19=0. Did you divide both sides by 4?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-3-which-of-the-follo...

We have the equation (4x^19)(2x + 9)(x - 3) = 0

This means 4x^19 = 0, (2x + 9) = 0 or (x - 3) = 0

Take 4x^19 = 0

Divide both sides by 4 to get: x^19 = 0

So, x = 0

Cheers,

Brent

## x² - 3x + 2 = 0

Factor to get: (x - 1)(x - 2) = 0

So, x = 1 or x = 2

Column A: Twice the sum of the roots of the equation

Sum of roots = 1 + 2 = 3

So, TWICE the sum = (2)(3) = 6

Column B: 6

How is sum roots= 1+2? and not root2+root1 ?

## You're referring to this

You're referring to this question: https://gre.myprepclub.com/forum/gre-math-challenge-33-x-2-3x-372.html

(Aside: in the future, please provide the link to the question so I know what you are referring to)

You are confusing ROOTS with SQUARE ROOTS

The "root" of an equation is the same as the "solution" to an equation.

For example, we can say:

- For the equation 2x = 10, the SOLUTION is x = 5

- For the equation 2x = 10, the ROOT is x = 5

Both of the above sentences have the same meaning.

Cheers,

Brent

## x(x−2)(x−3)(x−4)2=0x(x−2)(x−3

please explain

## Here's my full solution:

Here's my full solution: https://gre.myprepclub.com/forum/solve-x-x-2-x-3-x-2154.html#p30688

Cheers,

Brent

## Dear Brent,

I have a question. Is cross multiplication prohibited(as in dividing both sides of an equation by a variable is prohibited if it can be 0) under any condition? If yes, please explain a little bit.

Thanks in advance.

## Great question!

Great question!

I'm not sure if we're talking about the same thing when it comes to cross multiplication.

Here's how cross multiplication works:

If a/b = c/d, then we can say that ad = bc

This strategy will always work for two equivalent fractions, because the equation a/b = c/d indirectly tells us that a and d do NOT equal zero.

The reason for this is that a/0 does not equal a number. a/0 is undefined.

So, for example, saying that 6/0 = 3/0 makes no sense, since neither 6/0 nor 3/0 are actual numbers (in fact, saying 6/0 = 3/0 makes as much sense as saying that pencil = chicken :-)

Your other question: dividing both sides of an equation by a variable is prohibited if it can be 0

This is a different situation altogether.

If we have the equation: xy = xz, we CAN'T divide both sides by x to get y = z, since it COULD be the case that x = 0.

Does that help?

Cheers,

Brent

## Thanks, Brent. Your answer

I understand that both sides of an inequality cannot be multiplied by 0 since that can alter the relationship between the 2 sides(by making them both 0 in which case they will be forcefully equal).

Can both sides of an equation be multiplied by 0? I do not find any problem with that because the relationship between the 2 sides will still be the same as before.

Thanks.

## I suppose one COULD multiply

I suppose one COULD multiply both sides of an equation by zero, but the results aren't useful.

For example, say we want to solve the equation: 3x = 15

If we multiply both sides by zero, we get: 0 = 0

So, now what?

Cheers,

Brent

## Hi Brent,

https://gre.myprepclub.com/forum/if-3x-11278.html

(3x/4-2)^2 can be written as (3x-8/4)^2

now I had multiplied 4 so that denominator will get cancelled

so left wit 9x^2-64

(3x+4)(3x-4)

and i got x=4/3

and x=-4/3

Could you please suggest me where did I took a wrong step

Thanks

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-3x-11278.html

You're correct to say that: (3x/4 - 2)² = 0

Is the same as: [(3x - 8)/4]² = 0

Notice that [(3x - 8)/4]² = [(3x - 8)/4][(3x - 8)/4]

= (3x - 8)²/16 = 0

So we now have: (3x - 8)²/16 = 0

At this point we can multiply both sides by 16 to get: (3x - 8)² = 0

This tells us that 3x - 8 = 0

And this mean x = 8/3

ASIDE: Please note that (3x - 8)² does not equal 9x² - 64

(3x - 8)² = (3x - 8)(3x - 8)

= 9x² - 48x + 64

Cheers,

Brent

## But if we plug in x= +4 in 1

https://gre.myprepclub.com/forum/if-1-y-1-y-1-1-y-4-what-is-the-sum-of-all-solutions-for-y-21448.html

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-1-y-1-y-1-1-y-4-what-is-the-sum-of-a...

On that thread, I don't see anybody suggesting y = 4 is a possible solutions of the equation.

The two posters (including me) state that the solutions are y = 2 and y = -2.

## Hey Brent,

Can you please tell me should we use algebraic identity to solve this or square root (x + 20)^2 = 5

It's from GRE math review exercise 21 algebra part c......and if not algebraic than why not as it looks like it can be

## I would apply the property

I would apply the property that says:

If x² = k (where k ≥ 0), then either x = √k or x = -√k

Example: If x² = 9, then either x = 3 or x = -3

So for your particular question, we have: (x + 20)² = 5

So either x + 20 = √5 or x + 20 = -√5

If x + 20 = √5, then x = √5 - 20

If x + 20 = -√5, then x = -√5 - 20

## Hey Brent first of all bless

## I BELIEVE the only time I'd

I BELIEVE the only time I'd consider expanding would be if there were one or more variables on the other side of the equation.

Consider this equation: (x + 1)² = 3x + 7

In this case, due to the variable expression 3x on the right side, it wouldn't make any sense to write: EITHER x + 1 = √(3x + 7) OR x + 1 = -√(3x + 7)

However, if we expand we get: x² + 2x + 1 = 3x + 7

Set equal to zero: x² - x - 6 = 0

Factor: (x - 3)(x + 2) = 0

So, either x = 3 or x = -2