Algebra (all content)
- 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
- Solving simple quadratics review
- Solving quadratics by taking square roots: challenge
Sal discusses the exact order of steps in the process of solving the equation 3(x+6)^2=75. Created by Sal Khan.
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- how to find roots of a quadratic equation if b^2-4ac < 0 , if b^2-4ac is a perfect square i.e. (-64) .(2 votes)
- I understand the reason Sal is putting +/- in front of the square rooted number. However why is it that we do this only in quadratics and not other equations?(3 votes)
- Quadratics are not the only place where the +/- is used. We may be using functions like absolute value. But of the not-so-many places in math where the +/- will get used, quadratics is one of the places where it will be appearing most with little to no absence. We need it for absolute value so we can find which x-values, the positive and negative, will be adequate values for the overall function (and have outputs to the same p-value).
Most other places, however, do not need to use the +/- as a way to find the equation(s) and may only have at most one possible solution, not two.(2 votes)
- At3:10, why is it -6 plus or minus 5 instead of 5 plus or minus -6?(2 votes)
- You started with
x+6=±5, so subtracting 6 from both sides yields
If you started with
x+6=+5 or x+6=-5and subtracted 6 from both sides, you'd get
x=-6+5 or x=-6-5. And you could recombine those last two possibility back to just
x=-6±5. So it's the same thing either way: the
±sign is attached to the 5.(2 votes)
- on the options shown on the question,2:00Sal chooses a "take the square root of both sides" option. not to make myself dumb if it's too obvious, but i fail to see the difference between "square both sides" and "take the square root of both sides". HEELLP!(2 votes)
- They are opposite operations.
If you square 4, then you are doing 4^2 = 4*4 = 16
If you take the square root of 4, then you are doing sqrt(4) = 2
Do you see the difference?
If Sal had squared both sides of: (x+6)^2=25, here's what would have happened:
[(x+6)^2]^2 = 25^2
(x+6)^4 = 625
Squaring both sides just makes the equation even more complicated.
Hope this helps.(2 votes)
- It seems that the exercise for this video is missing. Can someone tell me where these exercises are located?(2 votes)
- It would be in "Algebra">"Quadratic Equations">"Solving Quadratics by Taking Square Root" and all the way at the bottom is "Understanding the Equation Solving Process" where a problem like this should appear.(2 votes)
- Do we have to write the plus or minus signs?(2 votes)
When you take the square root of both sides of an equation, you get two answers.
In the video, Sal had
(x+6)²=25 Take the square root of both sides
x+6 = ±√25
The plus or minus sign means you have two equations
x+6 = 5 and x+6 = -5
You don't have to write the ± sign but if you don't, you then need to write out both equations x+6 = 5 and x+6 = -5 and then solve each equation to get your two answers.
The ± sign is used to indicate that you have two equations you are solving at the same time.
I hope that helps make it click for you.(2 votes)
- Is it okay if we just put the steps and ignore figuring out the exact answer?(2 votes)
- How would you solve:
h(x) = -3(x-2)(x+2)
I am found the vertices x=2 and x= -2, but I am not sure how to find the axis of symmetry, AKA the maximum or minimum. PLEASE HELP!!(2 votes)
- to find the x-coordinate of the vertex (which is also the axis of symmetry):
1. first find the 2 x-intercepts(when y=0)-> (x,0) (x,0)
2.add the x values of the x-intercepts then divide them by 2 -> (x+x)/2
In your question the x-intercepts are (2,0)(-2,0)
(2+-2)/2 = 0/2 = 0
so the x-coordinate of the vertex and the axis of symmetry is 0
To find the max/min value we need to find the y- coordinate of the vertex
to do that we substitute the x coordinate of the vertex in the equation
so the vertex is (0,12)
and the max value(it has a maximum value because it opens down) is 12(1 vote)
- how can you solve:
1/2(x-1)^2 +5=23 can we solve it using this method(1 vote)
- Yes... you can.
1) subtract 5 from both sides: 1/2 (x-1)^2 = 18
2) Then multiply both sides by 2 to eliminate the fraction: (x-1)^2 = 36
3) Take square root of both sides: x - 1 = +/- sqrt(36) which simplifies to x - 1 = +/- 6
4) Add 1 to both sides: x = 1 +/- 6
So x = 7 and x = -5(2 votes)
- I don't believe it would change the answers, but would you not also have + or - (x+6) on the left side of the equation after taking the square root of both sides?(2 votes)
- You only need to have a plus or minus on one side. if you had it on both these would be the possible answers:
(x+6) = 25
(x+6) = -25
-(x+6) = 25
-(x+6) = -25
This is still only two possibilities though, because (x+6)=25 is the same as -(x+6)=-25, as are (x+6)=-25 and -(x+6)=25.
This is the same whenever you take the square root of both sides of an equation and use plus or minus, so you only need it on one side.(0 votes)
Use the cards below to create a list of steps in order that will solve the following equation. 3 times x plus 6 squared is equal to 75. And I encourage you to pause this video now and try to figure it out on your own. Figure out which of these steps and in what order you would do to solve for x here. So I'm assuming you've given it a go. So let's try to work through it together. And first, let me just rewrite the equation. So we have 3 times the quantity x plus 6 squared is equal to 75. So what I want to do is I want to isolate the x plus 6 squared on the left-hand side. Or another way of thinking about it-- I don't want this 3 here anymore. So how would I get rid of that 3? Well, I could divide the left-hand side by 3. But if I do that to only one side of the equation, it won't be equal anymore. These two things in yellow were equal to each other. If I want the equalities to hold, anything that I do to the left-hand side, I have to do the right-hand side. So let me divide that by 3 as well. And so on the left-hand side, I am left with x plus 6 squared is equal to 75 divided by 3. So 75 divided by 3 is 25. So actually, let me just pick out the first one I did. I divided both sides by 3. So that was my first step then. Let me write that in a darker color. So that was my first step right over there. Now let's think about what we're doing. We're saying that something squared is equal to 25. So this something could be the positive or negative square root of 25. So we could write this as x plus 6 is equal to the plus or minus square root of 25. So I'm essentially taking the positive and negative square root of both sides. So, let's see. This looks like this step. I took the square root of both sides. That's step number two. And so, let me just rewrite this. This is the same thing as x plus 6 is equal to plus or minus 5. And now I want to just have an x on the left-hand side. I want to solve for x. That's the goal from the beginning. So I would like to get rid of this 6. Well, the easiest way to do that is to subtract 6 from the left-hand side. But just like before, I can't just do it from one side of an equation. Then the equality wouldn't be true. We're literally saying that x plus 6 is equal to plus or minus 5. So x plus 6 minus 6 is going to be equal to plus or minus 5 minus 6. Or actually, let me write it this way. So let me subtract 6 from both sides. On the left-hand side, I'm left with an x. And on the right-hand side, I could write it this way. Let me do it in that green color. I have negative 6 plus or minus 5. So what are the possible values of x? Or actually, I keep forgetting. We don't have to actually give the value for x. We just have to say what steps we did. So then, let's see. After we took the square root of both sides, we then subtracted 6 from both sides. So that was step three right over there. Then that got us to essentially the two possible x's that would satisfy this equation right over here. And just for fun, let's actually solve it all the way. So if we solve it all the way, so x is equal to negative 6 plus 5 is negative 1, or x is equal to negative 6 minus 5 is negative 11. And you could verify that both of these work. If you put either of them in here-- if you put negative 1 here, you get negative 1 plus 6 squared is 5 squared. If you put negative 11 here, it's negative 11 plus 6 is negative 5 squared. Obviously either plus or minus 5 squared is going to be 25. 25 times 3 is 75. So these are our three steps. We divided both sides by 3. Then we took the square root of both sides. Then we subtracted 6 from both sides. And then we were essentially done. So let's input those steps. So the first thing we did, we divide both sides by 3. That's the first thing we did. And then we took the square root of both sides. And then we subtracted 6 from both sides. We got it right.