01 - Solving Equations in Quadratic Form - Part 1 (Learn to Solve Equations in Algebra)

In this lesson, you will learn what it means for an equation to be represented in "quadratic form". This means that the equation has the general form of a quadratic function, but the variables may be slightly more complex expressions.

To solve these types of equations, we first perform a substitution where we let the variable "z" represent a part of the equation. The result of this is a classic quadratic equation in "z" that can then be solved and represented in terms of the original variable "x".

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[khalilfuller]

5hr playlist. 1min down 4hrs & 59mins to go

[mathkh]

The math is really nice, and I hope we can connect it to the world of mathematics

[BlackStinga]

hey, really nice video I never knew this technique existed, I see the premise of what youve taught is breaking it down into quadratic form but one quick lil doubt, at 32:27 in the video the square root of 1 is -1 cuz -1 x -1 = 1. What you meant was if x^2 = -1, thats not possible but square root of x = -1 is possible, atleast I think

[davidbrathwaite]

The other reason there is only one true solution is that the highest power is one. So it still follows the rule that the number of solutions will be the value of the highest power.

[davidlocke]

When the solutiom is on the y-axis, is it open or closed?

[longtruong]

In the examples of quadratic equations, the 3rd equation when x=o, the function is undefined. So it can’t be a quadratic equation, is it?

[Vovwe-special]

Hey, pls I don't really understand the substitution method, I don't know if there's a topic before this one that has more detailed explanation about substitution than this, thank you

[ChristieNel]

sqrt(i^4)=-1, no? It appears to work in the equation, too.

[EcoFailure]

Thanks! But what grade are your videos?

[ChristieNel]

Coincidentally I was trying to fit a parabola through three points today. Initially I used a sigmoid function and could successfully calculate parameters to fit it through two points and could move the remaining parameters by hand to make it go through the third, but for the life of me can't solve the parameter for the third point mathematically. Hence I tried a parabola instead and ran into the same problem. Am I missing something or is this just not that trivial?

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