I've been watching your videos for quite some time now and honestly you bring my curiosity to the top. I've never been this excited about math. Thank you so much for your humble work and teaching the best for everyone here.
We always learn about how the discriminant works, but not the logic behind it. Thanks Eddie for making this clear. Just realised that Δ = 0 is equivalent to the axis of symmetry lol. Never thought of it like that.
I wish this was the quality at my high school. Rn in second year astronomy bachelor and hearing new things in this video or things that were only discussed in uni. But it's rlly cool to learn more about a concept that I've used a million times.
if the coefficient is not rational, then the rule "when ∆= perfect square, there are 2 rational solutions" doesnt apply? or the coefficients are assumed rational? eg. √2 x^2 + √24 x + √2 is a quadratic, and ∆ = 16 but the solutions are not rational, please could anyone save my confused mind
A football is being kicked through the uprights, along the path of y = -0.01*x^2-0.6*x where x and y indicate its position in yards, and where the kicker's position is at the origin. The bottom of the uprights is 3.33 yards above y=0, and the top of the uprights is 10 yards above y=0. Can this kick clear the field goal, and what range(s) of positions can the field goal be, relative to the placekicker in order to score? Solution: y = -0.01*x^2+0.6*x Set y=3.33, and solve for x: 3.33 = -0.01*x^2+0.6*x 0 = -0.01*x^2+0.6*x - 3.33 Check the discriminant. If it is less than zero, that means the parabola's vertex is too low to score. If it is greater than zero, that means there are two possible positions where the placekicker can be, in order to just barely clear the bottom of the field goal. If the discriminant is equal to zero, that means the football will just barely enter the uprights, but with only one option for where it can originate, due to the fact that it hits the uprights at its vertex. In this case, the coefficients on the parabola are a=-0.01, b=+0.6, and c=-3.33. The value of c will be 10, in the next trial after this. Trial 1 discriminant (bottom of the uprights): 0.6^2 - 4*(-0.01)*3.33 = 0.2268 Trial 2 discriminant (top of the uprights): 0.6^2 - 4*(-0.01)*10 = -0.04 The fact that Trial 2 has a negative discriminant, means that the football never rises above 9 yards. So our two solutions for x when y=3.33 will bound the range of values for x, where the placekicker can score. Continue solving the quadratic formula for a=-1, b=+0.6 and c=-3.33. We get x = 6.188 yards and x=53.81 yards. Anywhere within this range, the placekicker can score with a ball taking the given trajectory.
Has to be the best video of this quadratic theory stuff I could find thank you so much !! Explained much better than my teacher
I've been watching your videos for quite some time now and honestly you bring my curiosity to the top. I've never been this excited about math. Thank you so much for your humble work and teaching the best for everyone here.
Everytime i watch your videos, I learn something new
We always learn about how the discriminant works, but not the logic behind it. Thanks Eddie for making this clear. Just realised that Δ = 0 is equivalent to the axis of symmetry lol. Never thought of it like that.
Wonderful explanation! Thanks
I wish this was the quality at my high school. Rn in second year astronomy bachelor and hearing new things in this video or things that were only discussed in uni. But it's rlly cool to learn more about a concept that I've used a million times.
Two greatest teachers ON TH-cam
JONATHON OSBOURNE
EDDIE WOO
WHO IS JONATHON??????????????????????????????????????????????????????????
And Brian Maclogan
thanks, scout!
Very concise. Much appreciated sir.
if the coefficient is not rational, then the rule "when ∆= perfect square, there are 2 rational solutions" doesnt apply? or the coefficients are assumed rational?
eg. √2 x^2 + √24 x + √2 is a quadratic, and ∆ = 16 but the solutions are not rational, please could anyone save my confused mind
thank you so much!
Why is the descriminant called Delta, delta means difference?
Probably to reserve d to stand for coefficients in higher order polynomials.
Because it is the 'difference' in distance from the asymtote
Thank you so much for this!
that's the truth
Can you demonstrate it with practical problem?
A football is being kicked through the uprights, along the path of y = -0.01*x^2-0.6*x where x and y indicate its position in yards, and where the kicker's position is at the origin. The bottom of the uprights is 3.33 yards above y=0, and the top of the uprights is 10 yards above y=0. Can this kick clear the field goal, and what range(s) of positions can the field goal be, relative to the placekicker in order to score?
Solution:
y = -0.01*x^2+0.6*x
Set y=3.33, and solve for x:
3.33 = -0.01*x^2+0.6*x
0 = -0.01*x^2+0.6*x - 3.33
Check the discriminant. If it is less than zero, that means the parabola's vertex is too low to score. If it is greater than zero, that means there are two possible positions where the placekicker can be, in order to just barely clear the bottom of the field goal. If the discriminant is equal to zero, that means the football will just barely enter the uprights, but with only one option for where it can originate, due to the fact that it hits the uprights at its vertex.
In this case, the coefficients on the parabola are a=-0.01, b=+0.6, and c=-3.33. The value of c will be 10, in the next trial after this.
Trial 1 discriminant (bottom of the uprights): 0.6^2 - 4*(-0.01)*3.33 = 0.2268
Trial 2 discriminant (top of the uprights): 0.6^2 - 4*(-0.01)*10 = -0.04
The fact that Trial 2 has a negative discriminant, means that the football never rises above 9 yards. So our two solutions for x when y=3.33 will bound the range of values for x, where the placekicker can score.
Continue solving the quadratic formula for a=-1, b=+0.6 and c=-3.33. We get x = 6.188 yards and x=53.81 yards. Anywhere within this range, the placekicker can score with a ball taking the given trajectory.
Find the roots of the following quadratic: x^2+2x+2=0
"So helpful" is the least I can say :)
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Ohh sir love u i love u u r my best teacher pls pls support me i m very poor student pls help me sir