Highest and least value of 4x^2 in the the inequation
sry wait typing the question
I've seen all of these questions in my coaching module. o.O
Since No.Name isn't here, I'll type the question out...
\[\dfrac{6x^2 - 5x - 3}{x^2 - 2x + 6} \le 4\]
What did you get for the solution set of \(x\)?
My cputer hanged , sry yes that is the question
OK, your first step is to solve the inequality...
I am stuck on this step \[\huge \frac{ 2x^2 +3x-27 }{ x^2-2x+6 } <=0\]
They can't be perfectly factorized , o have I done a very silly mistake
The denominator is always positive, so you can just remove it from the inequality.
yeah \[\huge2x^2+3x-27 \le 0\]
Great, factorize.
one oot is in decimal;s
Yeah.\[2x^2 + 9x - 6x - 27 = x(2x + 9) - 3(2x + 9) = (x-3)(2x+9) \le 0\]\[x \in \left(-\frac{9}{2},3\right)\]
Now is the fun part.
Actually \[\huge [-4.5 , 3]\], they are included
Argh, OS is buggy. Yeah - I was correcting that.
but weird answer is given
No...
Do you know how \(|x|^2 = x^2\)?
yup
I am getting 36 and 81
So if a number has the highest modulus, then its square will be the highest. Which means that -4.5 has a bigger modulus than 3, so it has the highest square in the solution set of x, meaning that it has the highest four times the square...
Similarly the number with the least modulus in the solution set of x is 0, so it is supposed to have the least square, and also the least four-times-the-square.
You got the 81 right.
oh fu**k
You mean fu*k.
wtf man , I didn't realize that , I plugged in the exteme values lol
lol
0 is a cruel number thank you , !
Ancient Indians are coming to kill you. :P
rofl
0 :D is my favorite nothing :P
because you get it on your exam am I right?
xD lol
Nothing is your favourite yet something's you favourite because Hamiltonian of thoughts can never be 0 @ikram002p
just jokin, hehe
ik udit :P but still a nice joke
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