(x²-2x+ 1)(x+2)/(x²-5x+6)
> then x is
Answers
Answer:
x^2 - 5x + 6) /(x^2 + x +1) < 0 or
(x - 2)(x - 3)/(x^2 + x + 1) < 0 .···· ·· ··(1)
But the denominator(Dr) of the inequality is always positive for all real values of x because
Dr = x^2 + x + 1 = (x + (1/2))^2 + (3/4), also it is never zero, in the field of real numbers R but in the field of complex numbers C as (x^2 + x + 1) = (x - w)(x - w^2) where w is cube roots of unity . Therefore (1) ==> (x - 2)(x - 3) < 0 which in turn gives that 2 < x < 3 or x€ (2, 3) .
3-in-1 protection for moms-to-be helps protect baby too.
I assume the inequality to be solved as as shown.
x2−5x+6x2+x+1<0
The denominator is an always-positive function.
x2+x+1>0∀x∈R
Hence, only the numerator affects the sign of the function. The problem reduces to solving the following inequality.
x2−5x+6<0
⟹(x−2)(x−3)<0
⟹x∈(2,3)
To corroborate the result, here’s a graph.
x^2–5x+6/x^2+x+1 < 0
x^2 - 5x + 6 < 0
x^2 -2x-3x+6 < 0
x(x-2)-3(x-2) < 0
(x-2)(x-3) < 0
Therefore
2<x<3
If x² - 5x - 6 = 0 , what is x?
How do you solve the given inequality: x+2/1-x <1?
If x−1−−−−−√−x+1−−−−−√+1=0 , then 4x=?
If x is a real number and [x] is the greatest integer less than or equal to x, then 3|x|+2-[x] =0. Will the above equation have any real roots?
What is the value of x² - x?
Use the fact that the denominator is always >0 (prove it by completing square). Then you can multiply both parts by the denominator. Then solve
(x^2–5x+6)<0
(x-3)(x-2)<0
2<x<3
Amazon Prime Video - start a 30 day free trial.
LHS = ( (x-2)*(x-3) )
÷ ( (x + (1/2) ) ^2) + (3/4)
So Denominator is positive and minimum 3/4.
Hence (x-2)*(x-3) < 0 or negative.
So 2 < x < 3 Answer.
Answer:
hshrjdjjjjdjdjdjjdjhrhjsnjdjrnrjghrjiudyhjsishhwjjhs
hbfkgcjn
vfdhnd
hnsjrm
hdbnj
hrjr
hbstgh
hennejtj