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## Algebra 1 (Eureka Math/EngageNY)

### Unit 4: Lesson 7

Topic A: Lessons 6-7: Solving basic one-variable quadratic equations- Solving quadratics by taking square roots
- Solving quadratics by taking square roots
- Quadratics by taking square roots (intro)
- Solving quadratics by taking square roots examples
- Quadratics by taking square roots
- Solving quadratics by taking square roots: with steps
- Quadratics by taking square roots: strategy
- Quadratics by taking square roots: strategy
- Quadratics by taking square roots: with steps
- Quadratic equations word problem: triangle dimensions
- Quadratic equations word problem: box dimensions

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# Solving quadratics by taking square roots examples

CCSS.Math: ,

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?

- 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 ?(9 votes)- 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.(12 votes)

- So, would there be an actual way to figure out only one answer.2:21(6 votes)
- is it just me or i cant see to the other comments to this post?(0 votes)

- How are you supposed to solve equations that have square rooted prime numbers?

[*Example*: x + 8 = ^2](3 votes)- 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.(6 votes)

- it a parabola is open upwards what does it repersent? (in terms of a throwing arc)(5 votes)
- Not really. An upside down parabola could represent a ball being thrown, but not a bouncing ball, because it comes back up.(2 votes)

- 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...?(3 votes)- 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.(4 votes)

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

5 can't be square rooted.(1 vote)- 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.(5 votes)

- SO, since u got (x+3)^2=4. IS there's another way to do this ? I tried multiply ways

-U would go head and solve x+3^2 which could be 3x^2 or 9x=4 so that wont work

Also u can distributed x+3^2 to be 2x+9=4

-subtract by 9 which would be 2x=-5 that wont work either(1 vote)- 1) You can solve using square root method, as shown in the video.

2) You can simplify the equation and then factor. This is what you were trying to do, But, you are confusing what (x+3)^2 means. Here are the steps:

-- (x+3)^2 means (x+3)(x+3. Use FOIL, or distributive property to multiply it out.`(x+3)(x+3)=4`

`x^2 + 3x + 3x + 9 = 4`

`x^2 + 6x + 9 = 4`

-- Subtract 4 from both sides:`x^2 + 6x + 9 -4 = 4 - 4`

`x^2 + 6x + 5 = 0`

-- Factor the trinomial: find factors of 4 that also add to 6. Use 5 and 1`(x+5)(x+1) = 0`

-- Use the zero product rule. Split the factors and set each individually = 0 and solve`x + 5 = 0`

creates`x = -5`

`x + 1 = 0`

creates`x = -1`

You will later learn that you can also solve the equation by completing the square and by the quadratic formula.

Hope this helps.(4 votes)

- Forgive me if I'm wrong, but in another video, Sal said that a function can only have one x value for every y value. if you were to make a parabola on the y axis, would it not be a function? (sorry if this is confusing)(1 vote)
- Your wording/definition of a function is a litle off. If you have one x-value for every y-value, then you would have a vertical line, which is not a function.

A function is where each x-value creates exactly one y-value. A parabola is a function. If you turn the parabola sideways, then it is not a function.

Hope this helps.(4 votes)

- In the previous practice (find the zeros of: g(x) = -10x^2 +490) I ended up having the find the square root of -49, but I thought there wasn't a square root of -49 because there is no way to make it negative without cancelling it out... :/ I just ignored the negative sign and got the correct answer (positive or negative 7), but I don't see how that's possible.(1 vote)
- If you end up with 0 = -10x^2 + 490 to find the two solutions, you would divide by 10 to get 0 = - x^2 + 49, then adding x^2 to both sides gives x^2 = 49, so you are taking the square root of 49, not -49. Or if you divide by - 10, you end up with 0 = x^2 - 49 which is the difference of perfect squares and can be factored as 0 = (x+7)(x-7), so you still get ± 7.(4 votes)

- In a previous video concerning quadratics we did not include the integer preceding the quadratic expression in the equation to solve for the zero's of the function.

Example:

f(x)= 5(x-5)(x+1)

Step 1: 5(x-5)(x+1)=0

Step 2: (x-5=0) & (x+1=0)

The 5 is not included in the equation to solve for 0.

However, in this example we are including the subsequent integer.

Why?

What is the difference between the integer before the quadratic expression & after when it is included in a function?(1 vote)- The examples in this video are in vertex form: y=a(x-h)^2+k where (h,k) is the vertex of the parabola. The value of "a" in both examples is 1, so there is no lead coefficient like 5. The example you gave has a lead coefficient of 5, not 1. That 5 comes into play when you find any point other than the zeroes. It is a scalor that makes the parabola rise faster.

Does that help?(3 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.