Quadratic inequality

[Guest]

[tex]11x -7 < 2\sqrt(6x^2 - 22x + 20)[/tex]

I know how to solve a quadratic equation, but this inequality beat me.
I'm really not able to reach the solution, which is [tex](-\infty, 1)[/tex]
Thank you very much for your help.

[Guest]

my humble calculation gets......

97x^2 - 178x - 31 < 0

-0.16 < x > 2

.....................................

[Guest]

I always end up with with that solution, but wolfram alpha gives the correct one. And I don't know how to reach it.
Anyway, thank you.

[Guest]

Sorry, I always end up with:[tex]97x^2-66x-31<0[/tex]

[Guest]

The right hand side is only real when [tex]6x^2-22x+20\geq 0[/tex] which happens when [tex]x\leq5/3[/tex] or [tex]x\geq 2[/tex] so this is the domain we have to worry about. (Note that when the right hand side is real it is also non-negative because we are taking the positive square root.)

The square root in the inequality is hard to work with so it is tempting to get rid of it by squaring both sides. Unfortunately this doesn't work if one of the sides is negative ([tex]-3<1[/tex] is true but [tex](-3)^2<1^2[/tex] is false). However, we are allowed to square both sides if both sides are non-negative. So now there are two cases we have to consider:

Case 1: [tex]11x-7<0[/tex]
Which is true when [tex]x<7/11[/tex]. (The inequality we're interested in holds entirely in this range as the left hand side is negative, and the right hand side is a real non-negative number.)

Case 2: [tex]11x-7\geq 0[/tex]
This implies [tex]x\geq 7/11[/tex] and we are allowed to square the inequality to get rid of the square root to get [tex](11x-7)^2<4(6x^2-22x+20)[/tex]. This rearranges to [tex]97x^2-66x-31<0[/tex] which is true when [tex]-31/97<x<1[/tex]. Remember that [tex]x\geq 7/11[/tex] so the range this is valid is actually [tex]7/11\leq x<1[/tex].

Combining Cases 1 and 2, gives the answer: the inequality holds when [tex]x<7/11[/tex] or [tex]7/11\leq x<1[/tex]. Or more simply just [tex]x<1[/tex].

Hope this helped,

R. Baber.

[Guest]

I've just got it and written the answer to let others benefit from it.
Here it is:


Suppose we have this inequality: [tex]a<sqrt(b)[/tex].
To solve it, we must discuss all cases:

If [tex]a>=0[/tex], then[tex]b>a^2[/tex]
If[tex]a<0[/tex], then [tex]b>=0[/tex]

Now, coming to:[tex]11x-7<2\sqrt(6x^2-22x+20)[/tex]
1: if [tex]7x-11>= 0[/tex], then [tex]x>=7/11[/tex]. Its set is [tex][7/11,+\infty)[/tex]
and [tex](2\sqrt(6x^2-22x+20))^2 >(11x-7)^2[/tex]yields:
[tex]97x^2-66x-31<0[/tex].
Its set is [tex](-31/97,1)[/tex]

The intersection of the two sets gives: [tex][7/11,1)[/tex]

2: [tex]11x-7<0[/tex], then [tex]x< 7/11[/tex]. Its set is [tex](-\infty,7/11)[/tex]
and[tex]6x^2-22x+20>=0[/tex] is defined in [tex](-\infty, 5/3][/tex]
Their intersection is [tex](-\infty,7/11)[/tex]

Therefore the solution to the inequality is:

[tex][7/11,1) U (-\infty,7/11) =(-\infty, 1)[/tex]

Many, many thanks to anyone who helped.

[Guest]

R. Baber: Your answer helped a great deal. Thank you very much. I really appreciate. I'm looking forward to the day where I can help in my turn.