Solving a Radical Inequality

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#ChallengingMathProblems #RadicalInequalities

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[snejpu]

This is probably one of the easiest problems you've ever uploaded. Is it just a holiday atmosphere, or is it the silence before the storm? : )

[MathElite]

square both sides...
x>3/4
notice that inside the radical = (x-1)(x-3) >= 0
this tells us x is not in (1, 3)
x in (3/4, 1] U [3, infinity)

[justabunga]

Square both sides as x^2-4x+3<x^2. Notice that x^2 cancels each other out. All there is left is -4x+3<0, so x>-3/4. We have to be careful about the domain in the radical equation. The domain here is (0, 1] U [3, infinity). Putting these all together with the domain will be (-3/4, 1] U [3, infinity).

[amitsrivastava]

I always love the way you explain things and make them easy and interesting !!

[PunmasterSTP]

When I first saw that problem, my mind jumped immediately to squaring both sides. I think I would have examined the domain right after, but I like the sequence in which you did things. Thank you for sharing!

[angelmendez-rivera]

This seems like a single inequality, but actually, the inequality sqrt(x^2 – 4·x + 3) < x immediately implies 0 =< sqrt(x^2 – 4·x + 3), and importantly, 0 =< x^2 – 4·x + 3, and these follow immediately from the definition of sqrt as a function. The third inequality implies 0 =< x^2 – 4·x + 4 – 1 = (x – 2)^2 – 1, hence 1 =< (x – 2)^2, hence 1 =< |x – 2|. However, notice that the second inequality and the first inequality in connunction immediately imply 0 < x, so to solve 1 =< |x – 2|, we have 0 < x =< 1, and 3 =< x. With this restriction in mind, we have that x^2 – 4·x + 3 < x^2, equivalent to –4·x + 3 < 0, equivalent to 3 < 4·x, so 3/4 < x. Given the restrictions, this means 3/4 < x =< 1 or 3 =< x. This is the solution.

[theloganator]

Since visuals really are everything on YT, when you're on Desmos you can get your 3/4 label by clicking the wrench, then setting the x step size to 0.75. Ok, I lied, your tick label will be 0.75, but at least it will still label that and 3, since it's fortunately a multiple of 0.75.
Thanks for more great content!

[josephsilver]

A very nice problem. The solution set was discontinuous and thus is more of a challenge but your explanation was easy to understand. Good job!

[MyOneFiftiethOfADollar]

Conpleting square to get (x-2)^2-1 is easier way to see domain is (-inf, 1]. U [3, inf)

[seanfraser]

This is equivalent to x > 3/4 on the condition that x <=1 or x>=3. So the inequality holds on (3/4, 1]v[3, inf)

[carloshuertas]

Another great explanation, SyberMath!

[alexmorfi]

the explanation about x>0 is extra, you don't need to do it, just domain of radical is needed then its intersection by the last solution. by this maybe the audience asks you why you didn't say x must be >= than 0 because the radical of something could be 0.

[thefutureengineer]

x²-4x+3=(x-1)(x-3) and it must be positive so the domain of definition is
We have the inégalite
squere root x²-4x+3<x
Then x²-4x+3<x²
So we have - 4x+3<0
Finally we get x>(3/4) but from @ we say that x>3 is solution

[manojsurya]

This looks familiar after watchin lot of Ur radical equation videos, we just hafta check the domain n proceed.🌜🌝🌛

[italixgaming]

You compicated E-VE-RY-THING!!!!
We have sqrt(x²-4x+3)<x (1)
The LHS exists if and only if x²-4x+3 is positive.
But x²-4x+3=(x-1)(x-3)
The product of two numbers is positive if and only if both have the same sign. Then we exclude the region where x-1>0 and x-3<0 - we know that x-1>x-3 then we can't have x-3>0 and x-1<0 - which is ]1;3[. We don't need to study the function.

[Another way to see that is to transform x²-4x+3 into (x-2)²-1, and to say that (x-2)²-1<0 <=> |x-2|<1]

We also know that we can't have x<0 since the LHS is always positive (for the values of x where it is defined of course).
So (1) <=> sqrt(x²-4x+3)<|x| and x positive.
And when we elevate to the square we obtain:
(1) <=> x²-4x+3<x² and x²-4x+3 positive and x positive.
(1) <=> x>3/4 (which implies that x positive so we eliminate this condition) and x²-4x+3 positive (i.e. x is not part of ]1;3[.
(1) <=> x is element of ]3/4;1] U [3, +inf[.
We don't need any verification, since we kept the equivalence all the way.

[justabunga]

I figured that the graph here in the video is a rotated upper half of a hyperbola and should have 2 slanted asymptotes.

[agnibeshbasu]

Nice ur channel is growing so much! Keep growing

[rssl]

3-4x<0 then -4x<-3 then 4x>3 and x>3/4 is it correct ?
I did it by squaring and a bit of other stuff

[math]

very well done bro, thanks for sharing

[Ni]

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