Solving an Absolute Value Inequality with a Quadratic

[@ChrisHolly]

Boy, I wish I had you as a math teacher when I was a kid. I got about 98% of that, first time through, and I haven't had a math class since 1993.

[@AnishThe01]

Instead of that Test point method we here in India uses wavy curve method for most of our inequalities containing 0 on one side plot the points where the expression gets 0 (critical points) now the right most part of the no. line will give +ve values & sign of the expression changes (moving right to left) after every critical point whose factor have odd power & sign remains same for even powers. this method seems long in this comment but trust me it works magic in solving inequalities in seconds

[@devanshusoni9856]

Bhai class 11 mai hi sikha dete he or ye bhot easy question hai

[@sadidehsan6097]

An explanation video on why test point method works out would be great. I know why it works but would love to watch your explanation.

[@TheMathSorcerer]

Great suggestion!

[@k4rim799]

Same

[@kurtmueller2089]

I never heard of the term "test point method" I always called it the zero factor theorem, i.e. if a*b=0, then a=0, or b=0

[@nomicnevermic]

Since 8>0 this equation is possible for some x in R. Since we have an absolute value to be greater than a number we take x^2+6x+1>8 and x^2 +6x+1

[@nomicnevermic]

As for the method to find the sign of the quadratic (x-1)(x+7), we prefer to find the sign of each individual line (x-1 and x+7) and cross check them. In easy, 2nd degree quadratics since we know the general shape of such function we can also immediately say that the sign of coefficient a is the sign of the quadratic outside of the roots, with the opposite being the sign inside the roots. We are told to try test points if it's too difficult to determine the sign but the test is usually simple enough to be done in your head, as it doesn't look very scientific to write "yes!!!" or "no!!!" next to a wrong point test. We either write "impossible", "denied" (if possible) and "accepted"

[@marcushendriksen8415]

A quicker alternative to the test point method (when dealing with quadratic inequalities) is to take note of the sign on the x^2 term. If it's positive, then it's an upward-opening parabola and if it has distinct real roots r1 and r2, one can easily see that any x taken in the range of (r1, r2) will be evaluated to a negative number, no calculation required. A similar observation can be made when the x^2 term is negative.

[@83jbbentley]

Wizard, every night when my wife and child sleep I crank a few chapter-s a night. I’ve crossed over from College algebra to Trigonometry. I found considerable overlap from a CA-Trig to PreCal-Trig. CA teaches right angle Trig juxtaposed to Pre-Cal teaching unit circle. What is the reasoning for this?

[@TheMathSorcerer]

Yeah different books do it different ways:) That's the one of the things with math is that often things are introduced via different definitions.

[@83jbbentley]

@@TheMathSorcerer thank you Wizard.

[@acdude5266]

Great explanation and layout on the blackboard. ❤

[@prathiksaravanans2386]

I have a doubt, at one point you ended as (X+7)(X-1) >0 So, If we do as below👇, (X+7) >0, as the other term gets divided, And then we get, X > -7 So, in the solution you said that the solution set excluded the real values between -7 and 1. But as we get X > -7, we should exclude the numbers less than -7 right? I hope u got me. Pls clear my doubt

[@TheMathSorcerer]

(x + 7)(x - 1) > 0 implies both factors are positive or both factors are negative if they are both positive then x > -7 and x > 1, so x > 1 OR if they are both negative then x < -7 and x < 1, so x < -7, so we have x > 1 OR x < -7, so (-inf, -7) U (1, inf), this is another solution Note I didn't do it this way in the video, instead I used the test point method. But you can do it this way. It just becomes more complex when you have 3 and 4 factors. Hope this helps:)

[@prathiksaravanans2386]

@@TheMathSorcerer thank you very much. I got an idea

[@SuperYoonHo]

Thank you MS

[@theconservativeone2690]

Hi, can you please make a video on roadmap and books to learn pure mathematics from basic to advanced

[@asherm.pereira_796]

Yes please 🙏🤝

[@theoartis2076]

if you rewrite the question with an >= 8 you have to consider the point x={-3} because the f(-3)=8 and thats a local max of the function, -b/2a = -3 f(-b/2a)=8. This is really intresting because now the solutions add a only point

[@kummer45]

First of all when I meet these problems I ask what I am looking at. These are two functions that behaves in certain way where the constant function 8 is the ceiling that can never be reached. The graphs of these functions could give an idea of what is going on.

[@fernandocupil.6463]

Hey bro. Hace un momento estaba viendo un video tuyo sobre una demostración de un límite de una función cuadratica pero por la definición. Hya algo que no me queda muy claro y el el hecho de que Delta puede tomar valores, por decir un ejemplo(voy a escribir un ejemplo de los ejercicios que he estudiado) (1, E/3) (suponiendo que E signifique epsilón xD) Y me he quedado flipando porque no sé que significa. Ojala puedas explicarlo más adelante con oro video con más ejemplo de límites, me ayudaría mucho. Y noticias! Ya compr´mi libro Brand New de Calculus by Spivak! Lo estoy disfrutando mucho. Saludos and much love

[@shwez25]

if x is greater than -7 then how does it go to the left of -7 on the number line? Aren't -8, -9 , -10 smaller than -7? in which case shouldn't x be shaded to the right side of -7 ?

[@enocreyes7025]

Thank you.

[@aakh3500]

You do not need to check -8 and other cause we have x^2 that means parabolic branches are up

[@Iamblindanddeaf23]

Could you do absolute value inequality with irrational expression in rhs, where there is a parameter in it, perhaps pls

[@valajayrajsinh7532]

Thanks ☺️👍🥰

[@TheMathSorcerer]

You’re welcome 😊

[@valajayrajsinh7532]

@@TheMathSorcerer thanks friend 🥰👍😀

[@ฅคนเทรดเดอร์]

Thanks you

[@kpt123456]

good method alternatively both x+7 and x-1 are positive or both are -ve . then this becomes algebraic method

[@elitegorilla9440]

Test check is boring you could just get the factor behind x^2 change start from right side if postive start with + negative - .after root +,- change unless it's delta=0

[@昆仑云路]

The x^2+6x+9＜0 can have solutions like x＝-3+ki, here k is any real number and i＝√-1.

[@marcushendriksen8415]

Yes, but as he said, this is just basic algebra lol, so it's to be understood that we're only dealing with reals

[@pn4960]

I don’t understand why testing for 0 is enough and you don’t have to check for every integer between -7 and 1

[@aakh3500]

If a>0 in ax^2+bx+c=0 then parabolic branches are up and graph of it looks like U, so the middle part is under the x-axis

[@alexandroskourtis5268]

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