If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Solving quadratics using structure

Sal solves (2x-3)^2=4x-6 by substituting p for 2x-3 and obtaining the simpler equation p^2=2p.

## Want to join the conversation?

• at ( p^2 = 2p) why can't you divide both sides by p, giving p = 2 ?
• You can,but dividing by p will give you only one solution,that is p=2 but going through the "long process" will give you two answers p=2 and p=0,so it's not always advantageous to go for the shortcut.
• Is there a graphical/visual way to understand why:

(2x-3)^2 = 4x-6
(2x-3)^2 = 2(2x-3)
(2x-3)(2x-3) = 2(2x-3) i divide by 2x-3 from both sides
2x-3 = 2
2x -3 +3 = 2 + 3
2x = 5
2x/2 = 5/2
x = 2.5

eliminates one result? I kind of understand it, but not securely. If there was a way to visually understand this I think I'd feel more confident. Perhaps something related to the parable?
• Mhm, I'm not sure if a visualization is helpful here. The main point is, that the other result is `x = 1.5` and `2*(1.5) - 3 = 0`, so the division woudn't be possible at all. If you want to plot it: Plot `y=(2x-3)²`, `y=4x-6` and plot the term, you want to devide by (so `y=2x-3`). As `2x-3=0` can't be devided by, you "jump over" this point. But in the plot you can see, that exactly at this point your functions are intersecting.
In short: If you want to devide by some function (as `2x-3`), just plot it and look for the zeros. Check those points in addition.
• I don't understand how he factored out the p.
I guess I can understand how the exponent for the first p drops to one, but how does the second term turn into p-2??
• The 2 terms we're factoring are p^2 and -2p.
They each have a common factor of p,
so we remove the common factor to be outside the parentheses
and leave the other factors of p and - 2 inside the parentheses.
to get the expression p ( p - 2 ).
(As a check, do the multiplication to be sure we still have what we started with.)
• i did it this way..

(2x-3)^2=4x-6
(2x-3)^2=2(2x-3)
2x-3=2 : i divided both sides by 2x-3
2x=5
2x=5/3

is this legit??
how do i get the other x??
• This wouldnt work if the question is (2x-3)^2-12 = 4x-6. Even if you divide both sides with (2x-3) , it would become 2x-3-12/(2x-3)=2. Its much easier to just assume 2x-3=p from this start and solve for possible outcome for p then substitute back to 2x-3=p to find the possible x.
• ETA: I found the mistake - the very first step of expanding the parentheses, after doing it correctly everything worked perfectly using my usual method :D

I have several questions relating to the way I tried to solve it (which gave me a really counter-intuitive result):
1) are there any factors of a (positive) 60 whose sum would be -4? (the 60 is positive, the 4 is negative, specifically)

2) I suspect the answer to previous is "no"? Why is that? I thought a trinomial can always be factored using sum-product method?

3) also did I expand these parentheses correctly (2x -3)^2 --> 4x^2 +9

This is what I did:

Expand parentheses
4x^2 +9 = 4x -6

Get all of them on one side (-4x and +6 to both sides)
4x^2 -4x +15 = 0

Multiply coefficient on 1st term with that of the last term to get a number where to look for sum of factors
4*15 = 60

Which factors of 60 would add up to -4?
6*10=60 and 6 -10= -4 <-- Error, both of them need to be negative if I want their product to be a positive 60, this condition seems impossible, didn't notice at the time, counter-intuitively I still got both of Sal's results but also an additional one.

Can now split the middle term with my erroneous factors
4x^2 +6x -10x +15

Use grouping, extracting a factor from both sides
4x^2 +6x and -10x +15 give
2x(2x +3) -5(2x -3)

I have ended up with 3 binomials two of which is what Sal got, (2x-5) and (2x-3) and an additional one: (2x+3)
• For question 1: Solve by writing out a system of equations.
Statement 1: Factors of 60. Therefore, let the factors be x and y. You have xy = 60
Statement 2: Sum to -4. Therefore x + y = -4.

You now have two equations. Solve the quadratic formed:
Equation 1: xy = 60 -> x = 60/y
Equation 2: x+ y = -4 -> 60/y + y = -4
So you have:
60/y + y = -4
60 + y^2 = - 4y
y^2 + 4y + 60 = 0

Check to see if real roots are possible by using the discriminant (b^2-4ac). In this case, b^2 - 4ac is less than zero, which means that this system has no real roots. Therefore, no solution exists.

For question 2: I explained why no solution exists.
For question 3: I'm not quite sure I understand what the question was. If you are trying to expand (2x -3)^2, then your answer is incorrect. You're missing the middle factor (this is the one that always gets people).
(2x -3)^2
=(2x-3)(2x-3)
= 4x^2 -6x -6x + 9
= 4x^2 -12x + 9

Not sure if that answers your questions (particularly about Q3). If not, let me know.
• why didn't we divided both sides by p_ since _p^2 is
p * p like follow p^2/p = 2p/p >>> p = 2 and if this can't be then why did Sal factor out p_ as follow
_p^2 - 2p = 0 >>> p(p - 2) = 0
and as far as I know factoring out is dividing the terms by a factor which all the terms I want to divide accept division by
• Quadratic equations will have 2 solutions unless the 2 solutions happen to be the same, then it degrades to 1 solution. By dividing by "p", you destroy/lose one of the two solutions.

Basic rule: Never divide an equation by the variable or something containing the variable. You can lose potential solutions to the equation.

Hope this helps.
• Why are the questions in the videos super easy, then the questions they give you on the test like 10 times harder?? I can't figure this out for the life of me. Can someone pls help me?
• p-2=0 is not p=2, it would be p= -2(right?)
• No, sorry... p-2 = 0 becomes p = 2
Remember, you have to do the opposite operation to move the 2 to the other side of the equals. Since it is "-2", you need to use "+2" on both sides to move it. This creates p = 2.
• at when p^2 = 2p, why wouldn't you solve it as sqrt(p^2) = sqrt(2p)
• You could do it that way, you'd get p = √(2p). The solution would be the same (0 = √0, 2 = √4). It's just easier the way Sal does it, p(p-2) = 0, where you can clearly see the solutions are 0/2.
• Is there another way to do this problem. I know that there is but I don't know what the other way would be. Can someone PLEASE help me and coment down a faster/easier/different way of doing this problem.
• Sal's method is the fastest way.

The alternative way would be to expand the multiplication on left side of the equation.
(2x-3)^2 = 4x-6
4x^2-12x+9 = 4x-6
Subtract 4x and add 6 to both sides
4x^2-16x+15 = 0
Then, factor the trinomial using grouping.
AC = 60. Find factors of 60 that add to -16.
The factors would by -6 and -10
4x^2-6x-10x+15 = 0
2x(2x-3)-5(2x-3) = 0
(2x-3)(2x-5) = 0
Use zero product rule
2x-3= 0 and 2x-5=0
Solve each and you get the solutions
x = 3/2 or 1.5
x= 5/2 or 2.5

Hope this helps.