Yes, x-intercepts to be specific. The x-intercepts of a function are the points where the graph of the function crosses the x-axis. However, that is not the full story, roots of a function or an equation are values of x for which the function equals zero, this is what I mean, here is a function, f(x)=x²-1. Now lets make the function zero, we now have x²-1=0, let's solve this equation for x( solving also means finding the solution(s) to the equation),I am assuming you know factorization, (x-1)(x+1)=0, so then there is two solutions, x=-1 or x=1. You can test it yourself, if you make x=1 or x=-1 and put it in the equation x²-1, you will get 0, so indeed, -1 and 1 are the solutions to the equation. Now if you were to draw this function, you will see that at x=-1 and x=1 the graph crosses the x axis. So lets recap, roots are x-intercepts of the function, but roots are also the solutions to the equation f(x)=0, in the example above f(x) is x²-1, so while roots and x-intercepts refer to the same numerical values (that is., they are both the solutions to the equation 𝑓(x)=0, "roots" emphasizes the solution to the equation, and "x-intercepts" refers more to the graphical representation where these solutions cause the graph to intersect the x-axis. I know its a lot to take in, but I hope it helps. After reading all this, watch this video again, to see if it makes more sense.
If the discriminant is lesser than 0 then it has to be a negative number and the sqrt of a negative number is always a multiple of 'i' [ i=sqrt (-1) ] and i is not a real number. Hence If the discriminant is lesser than 0 then the root is a non-real root.
I’m 12 and your videos on quadratics are SO HELPFULL
common W
How's being 15
The date on the board being august 3, 2015 is kinda crazy. Like these kids r probably a couple years out of college now
Why is triangle used for discriminant I thought the triangle means change
the triangle is the Greek letter delta which stands for 'd' or 'discriminant'
it's the "change" to x away from the axis of symmetry x = -b/2a you would have to make in order to reach either side of the parabola.
Thank you soooooo much!
what does he mean by roots? intercepts?
Yes, x-intercepts to be specific. The x-intercepts of a function are the points where the graph of the function crosses the x-axis. However, that is not the full story, roots of a function or an equation are values of x for which the function equals zero, this is what I mean, here is a function, f(x)=x²-1. Now lets make the function zero, we now have x²-1=0, let's solve this equation for x( solving also means finding the solution(s) to the equation),I am assuming you know factorization, (x-1)(x+1)=0, so then there is two solutions, x=-1 or x=1. You can test it yourself, if you make x=1 or x=-1 and put it in the equation x²-1, you will get 0, so indeed, -1 and 1 are the solutions to the equation. Now if you were to draw this function, you will see that at x=-1 and x=1 the graph crosses the x axis. So lets recap, roots are x-intercepts of the function, but roots are also the solutions to the equation f(x)=0, in the example above f(x) is x²-1, so while roots and x-intercepts refer to the same numerical values (that is., they are both the solutions to the equation 𝑓(x)=0, "roots" emphasizes the solution to the equation, and "x-intercepts" refers more to the graphical representation where these solutions cause the graph to intersect the x-axis. I know its a lot to take in, but I hope it helps. After reading all this, watch this video again, to see if it makes more sense.
Im writing this comment a day before it be exactly uploaded for 8 years
4:41 TELL ME THE REASON WHY!!!!
If the discriminant is lesser than 0 then it has to be a negative number and the sqrt of a negative number is always a multiple of 'i' [ i=sqrt (-1) ] and i is not a real number. Hence If the discriminant is lesser than 0 then the root is a non-real root.
The girl who shouted imaginary,
She's in many videos
And I think she's a genius!!!😍
@@phenomenalphysics3548 are you simping over a disembodied voice?
I don’t think she’s real.
she's 'unreal'@@Steve.M
two solutions