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## Algebra 1

### Course: Algebra 1 > Unit 14

Lesson 5: Solving quadratics by factoring- Solving quadratics by factoring
- Solving quadratics by factoring
- Quadratics by factoring (intro)
- Solving quadratics by factoring: leading coefficient ≠ 1
- Quadratics by factoring
- Solving quadratics using structure
- Solve equations using structure
- Quadratic equations word problem: triangle dimensions
- Quadratic equations word problem: box dimensions
- Solving quadratics by factoring review

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# 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?

- at1:50( p^2 = 2p) why can't you divide both sides by p, giving p = 2 ?(113 votes)
- 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.(22 votes)

- 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?(25 votes)- 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.(11 votes)

- 2:06I 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??(15 votes)- 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.)(13 votes)

- 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??(13 votes)- 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.(1 vote)

**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)(12 votes)- 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.(7 votes)

- p-2=0 is not p=2, it would be p= -2(right?)(5 votes)
- 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.(13 votes)

- at2:04when p^2 = 2p, why wouldn't you solve it as sqrt(p^2) = sqrt(2p)(5 votes)
- 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.(5 votes)

- I'm stuck on the p part can someone please help me. Greatly appreciated.

Owo(4 votes)- This might be a couple of days late but here we go...

The whole point of using p is to make the existing equation into a much simpler one. Using p allows you to find x easier. Once you substitute p into all the places possible, then you can simplify and use the zero product property to find what the solutions are. In this case, p=0 and =2. You can then make p=2x-3. You substitute p for one of the solutions of p(0 or 2). You can then simplify to find the value of x.

I really hope this helps!(4 votes)

- Is there any other way/s of solving the equation in this example?(5 votes)
- 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.(1 vote)- 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.(7 votes)

## Video transcript

- [Voiceover] So let's
try to find the solutions to this equation right over here. We have the quantity two
X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind
of a classic quadratic form, but there might be a faster
or a simpler way to do this if you really pay
attention to the structure of both sides of this equation. Well let's look at this, we
have two X minus three squared on the left-hand side, on the right-hand side
we have four X minus six. Well four X minus six,
that's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. And so this is really interesting, we have something squared is equal to two times that something. So if we can solve for the something, let me be very clear here, so the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for X, and I'll show you that right now. So let's say, let's just
replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So let's say that P is equal to two X minus three. Well then this equation
simplifies quite nicely, the left-hand side becomes P squared, P squared is equal to two times P, 'cause once again two X minus three is P, two times P. And now we just have to solve for P. And I'll switch to just one color now. So we can write this as, if we subtract two P from both sides, we can get P squared minus two P is equal to zero, and we can factor out a P, so we get P times P minus two is equal to zero. And we've seen this shown multiple times, if I have the product of two
things and they equal to zero, at least one of them
needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. Well if P minus two is equal to zero, then that means P is equal to two. So either P equals zero, or P equals two. Well we're not quite done yet, because we wanted to solve for X, and not for P. But luckily we know that two X minus three is equal to P. So now we could say either two X minus three is going
to be equal to this P value, is going to be equal to zero, or two X minus three is going
to be equal to this P value, is going to be equal to two. And so this is pretty
straightforward to solve, add three to both sides, you get two X is equal to three. Divide both sides by two, and we get X is equal to 3/2, or over here if we add
three to both sides, we get two X is equal to five. Divide both sides by two, and you get X is equal to 5/2. So these are the possible solutions, and this is pretty neat. This one right over here, you could almost do this in your head, it was nice and simple. Well if you were to expand this out, and then subtract this, it would have been a much more complex set of operations that you would have done. You still would have hopefully
gotten to the right answer, but it would have just
taken a lot more steps. But here we could appreciate some patterns that we saw in our equations, and namely, we have
this thing being squared and then we have two
times that same thing, two times two X minus three.