Question 1039273: HELP!!! Determine the values of the variable for which the expression is defined as a real number(answer in interval notation)?
1. Sqrt of 5(x)^2 -7x+2
2. (x)^5 +(x)^3 greater or equal to (x)^2 +5x. (Interval notation rounded to two decimals).
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! the first is where 5x^2-7x+2>=0
(5x-2)(x-1)> = 0
the roots are x=(2/5) and x=1
Between these, the value is <0
so the intervals are(-oo,2/5] and [1,oo)
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x^5+x^3>=x^2+5x
x^5+x^3-x^2-5x>=0.
This is <0 when x <-1.23
It is >=0 when x is between -1.23 and 0
It is <0 when x is between (0,1.44)
It is >0 when x>=1.44
[-1.23,0]U[1.44,oo]
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Determine the values of the variable for which the expression is defined as a real number(answer in interval notation)?
1. Sqrt of 5(x)^2 -7x+2
Solve:: 5x^2-7x+2 >= 0
(5x-2)(x-1) >= 0
Draw a number line and plot x = 2/5 and x = 1
Find the solution intervals::
For x <= 2/5:: -*- > 0 ; true ; solutions in (-oo,2/5]
for 2/5< x <1:: +*- > 0; false ; no solutions in (2/5,1)
for 1 <= x <+oo:: +*+ >0 true ; solutions in [1,+oo)
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2. (x)^5 +(x)^3 greater or equal to (x)^2 +5x. (Interval notation rounded to two decimal places).
x^5+x^3-x^2-5x >= 0
x[x^4+x^2-x-5] >= 0
Solve x^4+x^2-x-5 >=0
x = -1.23 or x = 1.44
Solutions::
For -oo 0 ; false ; no solutions in (-oo,-1.23)
For -1.23<= x <0 ;; -*- > 0 ; true ; solutions in [-1.23,0]
For 0< x < 1.44 :: +*- > 0 ; false ; no solutions in (0,1.44)
For 1.44<= x <+oo :: +*+ > 0 ; true ; solutions in [1.44,+oo
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Cheers,
Stan H.
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