proof the inequality
- Thread starter Lizzy1
- Start date
I need some help at this inequality.. I have to proof that x^4-x^3+4x^2+3x+5>0 ; x is a real number
have you tried anything yet? can you post what you've tried?
I've tried to write it as some perfect squares , but it didn't work... one of my friends had to do this too , and he told me that , in the end , he got : x^4+x^2*(x-1)^2+6x^2+(x+3)^2+1>0 , but I don't know how to get there
.. (I know that that is true , because a a perfect square is bigger or equal with 0 , so if you add 1 , the sum will be bigger than 0)
D
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I've tried to write it as some perfect squares , but it didn't work... one of my friends had to do this too , and he told me that , in the end , he got : x^4+x^2*(x-1)^2+6x^2+(x+3)^2+1>0 , but I don't know how to get there
You are on correct thought process. There are several ways to prove this. One of the ways would be:
x^4-x^3+4x^2+3x+5
= x^4 - x^3 + 4* [x^2 + 2*3/8 * x + (3/8)^2] + 5 - 4*(3/8)^2
= x^2 * (x^2 - x) + 4* [x + (3/8)]^2 + 5 - 9/16
Now continue.....
Thank you , but I have a question:-D... why 3/8? ( and at the end , after +5 shouldn't be -4*9/16? .. anyway I'll try to continue it
)What SK did was this:Thank you , but I have a question:-D... why 3/8? ( and at the end , after +5 shouldn't be -4*9/16? .. anyway I'll try to continue it
\(\displaystyle x^4 - x^3 + 4x^2 + 3x + 5 = \)
\(\displaystyle x^4 - x^3 + (4x^2 + 3x) + 5 = \)
\(\displaystyle x^4 - x^3 + 4\left(x^2 + \dfrac{3}{4} * x\right) + 5 = \)
\(\displaystyle x^4 - x^3 + 4\left(x^2 + 2 * \dfrac{3}{8} * x\right) + 5 = \)
\(\displaystyle x^4 - x^3 + 4\left\{x^2 + 2 * \dfrac{3}{8} * x + \left(\dfrac{3}{8}\right)^2\right\} + 5 - 4\left(\dfrac{3}{8}\right)^2 = \)
\(\displaystyle x^2(x^2 - x) + 4\left(x + \dfrac{3}{8}\right)^2 + 5 - 4\left(\dfrac{9}{64}\right) = \)
\(\displaystyle x^2(x^2 - x) + 4\left(x + \dfrac{3}{8}\right)^2 + 5 - \dfrac{9}{16}.\)
Thank you