kindly help i this confusion (x+7)^2-49=0 (x+7)=+/-7 x=0 or -14 BUT COULDN'T I DO THE FOLLOWING (x+7)^2-49=0 x^2+14x+49-49=0 x^2+14x=0, then x^2=-14x , divide both sides by x x=-14 in this i cant discover x=0, how this could happen?! how to follow valid steps and miss one of the solutions ?

Never divide by the variable. You destroy / lose solutions to the equation. Instead, use factoring: x^2+14x=0 becomes x(x+14)=0 Use zero product rule: x=0 and x+14=0 Solutions become: x=0 and x=-14 Hope this helps.

it a parabola is open upwards what does it repersent? (in terms of a throwing arc)

I just realized that it could represent a ball being thrown into water because it could come back up.

How are you supposed to solve equations that have square rooted prime numbers? [_Example_: x + 8 = ^2]

You cannot combine numbers and roots, so by subtracting 8, the answer would be x = - 8 + √2. You can get an approximation if you need to, but the exact answer is as above.

how f(x)=(x-2)^2-9 is a function ? as sal said that functions can only have one output then how can a parabola be a function...?

A function generates only one output (y-value) for each input value (x-value). The equation f(x)=(x-2)^2-9 satisfies that requirement. Each time you input a value for "x", you only get one value for "y". I think you may be confusing "x" and "y". Parabolas will have the same "y" value for 2 different "x" values. That's ok. "y" is not the input.

Wouldn't it be 'and'? x=-1 and x=-5. My reasoning: It is a parabola and there are two x intercepts.

I'm not sure what your reasoning mean, since if its an actual reasoning, it will also apply to using 'or'. x = -1 or x = -5 means the x-value when f(x) = 0. You can't have a point with 2 x-values. You use x = -1 and x = -5 when you are describing the parabola, so you say f(x) has 2 x-intercepts, x = -1 and x = -5.

how do i add a square root symbol

On KhanAcademy, if your answer needs to include a square root, then a menu of special symbols is provided when your cursor is in the answer box. If no option is provided, then likely you haven't simplified your answer. For example: sqrt(49) would not be an appropriate answer because it simplifies to 7. Hope this helps.

at 4:00 the problem has (x-2)^2 .... When I tried I thought I couldnt take sqrt of negative numbers? edit: (Ah it's because the exponent is already there to save them from the sqrt cage) So it became: x^2 + 4 = 9 x = sqrt(5) why I couldnt get the answer this way?

There are several issues with your approach. Issue 1) (x-2)^2 does not equal x^2+4. Squaring a binomial always creates a trinomial. (x-2)^2 = x^2-2x-2x+4 = x^2-4x+4 Issue 2) When you applied the square root to: x^2=5, you lost one of the 2 solutions. You should have gotten x=sqrt(5) and x=-sqrt(5). The correct way to approach the problem is to recognize (x-2)^2 as a perfect square. Thus, sqrt[(x-2)^2] = x-2 (as shown in the video). Hope this helps.

what if it can't be square rooted? like ((x−6)^2)−5=0 5 can't be square rooted.

The 5 can be square rooted. It is just an irrational number which we leave as sqrt(5) unless we need an estimated value. Then we would use a calculator to get the estimate. You equation has solutions of: x = 6 +/- sqrt(5) Hope this helps.

Main content

Solving quadratics by taking square roots examples

Name: Solving quadratics by taking square roots examples
Uploaded: 2016-03-17T03:15:03Z
Description: Sal solves the equation (x+3)²-4=0 and finds the x-intercepts of f(x)=(x-2)²-9.

Google Classroom

Sal solves the equation (x+3)²-4=0 and finds the x-intercepts of f(x)=(x-2)²-9.

Want to join the conversation?

Sort by:

Ahmed Nasret
Posted 5 years ago. Direct link to Ahmed Nasret's post “kindly help i this confus...”
kindly help i this confusion
(x+7)^2-49=0
(x+7)=+/-7
x=0 or -14
BUT COULDN'T I DO THE FOLLOWING
(x+7)^2-49=0
x^2+14x+49-49=0
x^2+14x=0, then x^2=-14x , divide both sides by x
x=-14
in this i cant discover x=0, how this could happen?! how to follow valid steps and miss one of the solutions ?
Button navigates to signup pageComment on Ahmed Nasret's post “kindly help i this confus...”
(14 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “Never divide by the varia...”
  Never divide by the variable. You destroy / lose solutions to the equation.
  Instead, use factoring: x^2+14x=0 becomes x(x+14)=0
  Use zero product rule: x=0 and x+14=0
  Solutions become: x=0 and x=-14
  
  Hope this helps.
  Comment on Kim Seidel's post “Never divide by the varia...”
  (31 votes)
Ghostrider17784
Posted 6 years ago. Direct link to Ghostrider17784's post “it a parabola is open upw...”
it a parabola is open upwards what does it repersent? (in terms of a throwing arc)
Button navigates to signup pageButton navigates to signup page
(9 votes)
Answer
- Ghostrider17784
  Posted 6 years ago. Direct link to Ghostrider17784's post “I just realized that it c...”
  I just realized that it could represent a ball being thrown into water because it could come back up.
  Comment on Ghostrider17784's post “I just realized that it c...”
  (22 votes)
Br Paul
Posted 5 years ago. Direct link to Br Paul's post “How are you supposed to s...”
How are you supposed to solve equations that have square rooted prime numbers?

[Example: x + 8 = ^2]
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “You cannot combine number...”
  You cannot combine numbers and roots, so by subtracting 8, the answer would be x = - 8 + √2. You can get an approximation if you need to, but the exact answer is as above.
  Button navigates to signup page
  (8 votes)
Crafter
Posted 8 years ago. Direct link to Crafter's post “So, would there be an act...”
So, would there be an actual way to figure out only one answer.
2:21
Button navigates to signup pageComment on Crafter's post “So, would there be an act...”
(6 votes)
Answer
- Kura
  Posted a year ago. Direct link to Kura's post “you can not figure out on...”
  you can not figure out only one answer because it has 2 answers.
  Button navigates to signup page
  (1 vote)
TheCoolDitto91
Posted a year ago. Direct link to TheCoolDitto91's post “Do you think he started t...”
Do you think he started the video a little too late? He doesn't give his usual intro in this video.
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
Kishan Rusia
Posted 7 years ago. Direct link to Kishan Rusia's post “how f(x)=(x-2)^2-9 is a ...”
how f(x)=(x-2)^2-9 is a function ?
as sal said that functions can only have one output then how can a parabola be a function...?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “A function generates only...”
  A function generates only one output (y-value) for each input value (x-value).
  The equation f(x)=(x-2)^2-9 satisfies that requirement. Each time you input a value for "x", you only get one value for "y".
  
  I think you may be confusing "x" and "y". Parabolas will have the same "y" value for 2 different "x" values. That's ok. "y" is not the input.
  Comment on Kim Seidel's post “A function generates only...”
  (4 votes)
bre
Posted a year ago. Direct link to bre's post “how do i add a square roo...”
how do i add a square root symbol
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “On KhanAcademy, if your a...”
  On KhanAcademy, if your answer needs to include a square root, then a menu of special symbols is provided when your cursor is in the answer box. If no option is provided, then likely you haven't simplified your answer. For example: sqrt(49) would not be an appropriate answer because it simplifies to 7.
  
  Hope this helps.
  Comment on Kim Seidel's post “On KhanAcademy, if your a...”
  (6 votes)
malenia, blade of miquella
Posted a month ago. Direct link to malenia, blade of miquella's post “at 4:00 the problem has (...”
at
4:00
the problem has (x-2)^2 .... When I tried I thought I couldnt take sqrt of negative numbers?
edit: (Ah it's because the exponent is already there to save them from the sqrt cage)
So it became:
x^2 + 4 = 9
x = sqrt(5)
why I couldnt get the answer this way?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted a month ago. Direct link to Kim Seidel's post “There are several issues ...”
  There are several issues with your approach.
  
  Issue 1) (x-2)^2 does not equal x^2+4. Squaring a binomial always creates a trinomial. (x-2)^2 = x^2-2x-2x+4 = x^2-4x+4
  
  Issue 2) When you applied the square root to: x^2=5, you lost one of the 2 solutions. You should have gotten x=sqrt(5) and x=-sqrt(5).
  
  The correct way to approach the problem is to recognize (x-2)^2 as a perfect square. Thus, sqrt[(x-2)^2] = x-2 (as shown in the video).
  
  Hope this helps.
  Button navigates to signup page
  (4 votes)
williamxelder
Posted a year ago. Direct link to williamxelder's post “Wouldn't it be 'and'? x=-...”
Wouldn't it be 'and'? x=-1 and x=-5. My reasoning: It is a parabola and there are two x intercepts.
Button navigates to signup pageComment on williamxelder's post “Wouldn't it be 'and'? x=-...”
(3 votes)
Answer
- joshua
  Posted a year ago. Direct link to joshua's post “I'm not sure what your re...”
  I'm not sure what your reasoning mean, since if its an actual reasoning, it will also apply to using 'or'.
  
  x = -1 or x = -5 means the x-value when f(x) = 0. You can't have a point with 2 x-values.
  You use x = -1 and x = -5 when you are describing the parabola, so you say
  f(x) has 2 x-intercepts, x = -1 and x = -5.
  Button navigates to signup page
  (2 votes)
Name
Posted 3 years ago. Direct link to Name's post “what if it can't be squar...”
what if it can't be square rooted? like ((x−6)^2)−5=0
5 can't be square rooted.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 3 years ago. Direct link to Kim Seidel's post “The 5 can be square roote...”
  The 5 can be square rooted. It is just an irrational number which we leave as sqrt(5) unless we need an estimated value. Then we would use a calculator to get the estimate.
  You equation has solutions of: x = 6 +/- sqrt(5)
  
  Hope this helps.
  Comment on Kim Seidel's post “The 5 can be square roote...”
  (5 votes)

Video transcript

- [Voiceover] So pause the video and see if you can solve for x here. Figure out which x-values will satisfy this equation. All right, let's work through this. So, the way I'm gonna do this is I'm gonna isolate the x plus three squared on one side and the best way to do that is to add four to both sides. So, adding four to both sides will get rid of this four, subtracting four, this negative four on the left-hand side. And so we're just left with x plus three squared. X plus three squared. And on the right-hand side I'm just gonna have zero plus four. So, x plus three squared is equal to four. And so now, I could take the square root of both sides and, or, another way of thinking about it, if I have something-squared equaling four, I could say that that something needs to either be positive or negative two. So, one way of thinking about it is, I'm saying that x plus three is going to be equal to the plus or minus square root of that four. And hopefully this makes intuitive sense for you. If something-squared is equal to four, that means that the something, that means that this something right over here, is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two-squared is equal to four. If x plus three was negative two, negative two-squared is equal to four. So, either of these would satisfy our equation. So, if x plus three is equal to two, we could just subtract three from both sides to solve for x and we're left with x is equal to negative one. Or, over here we could subtract three from both sides to solve for x. So, or, x is equal to negative two minus three is negative five. So, those are the two possible solutions and you can verify that. Take these x-values, substitute it back in, and then you can see when you substitute it back in if you substitute x equals negative one, then x plus three is equal to two, two-squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two, squared is positive four, minus four is also equal to zero. So, these are the two possible x-values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way. So, we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x-values does the graph of y equals f of x intersect the x-axis. So, if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function. If I have the graph of some function that looks something like that. Let's say that the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x-values where you intersect, where you intersect the x-axis. Well, in order to intersect the x-axis, y must be equal to zero. So, y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So, figuring out the x-values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying, "For what x-values does f of x equal zero?" So we could just say, "For what x-values does this thing right over here "equal zero?" So, let me just write that down. So we could rewrite this as x, x minus two squared minus nine equals zero. We could add nine to both sides and so we could get x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So, we could say x minus two is equal to positive three or x minus two is equal to negative three. Well, you add two to both sides of this, you get x is equal to five, or x is equal to, if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that. If x is equal to five, five minus two is three, squared is nine, minus nine is zero. So, the point five comma zero is going to be on this graph. And also, if x is equal to negative one, negative one minus two, negative three. Squared is positive nine, minus nine is zero. So, also the point negative one comma zero is on this graph. So those are the points where, those are the x-values where the function intersects the x-axis.

Algebra (all content)

Course: Algebra (all content) > Unit 9

Want to join the conversation?

Video transcript