proof the inequality

[Lizzy1]

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

 

[Romsek]

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?

 

[Lizzy1]

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)

 

[Deleted member 4993]

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)

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.....

 

[Lizzy1]

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)

 

[JeffM]

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:

$\displaystyle x^4 - x^3 + 4x^2 + 3x + 5 =$

$\displaystyle x^4 - x^3 + (4x^2 + 3x) + 5 =$

$\displaystyle x^4 - x^3 + 4\left(x^2 + \dfrac{3}{4} * x\right) + 5 =$

$\displaystyle x^4 - x^3 + 4\left(x^2 + 2 * \dfrac{3}{8} * x\right) + 5 =$

$\displaystyle x^4 - x^3 + 4\left\{x^2 + 2 * \dfrac{3}{8} * x + \left(\dfrac{3}{8}\right)^2\right\} + 5 - 4\left(\dfrac{3}{8}\right)^2 =$

$\displaystyle x^2(x^2 - x) + 4\left(x + \dfrac{3}{8}\right)^2 + 5 - 4\left(\dfrac{9}{64}\right) =$

$\displaystyle x^2(x^2 - x) + 4\left(x + \dfrac{3}{8}\right)^2 + 5 - \dfrac{9}{16}.$

 

[Lizzy1]

What SK did was this:

$\displaystyle x^4 - x^3 + 4x^2 + 3x + 5 =$

$\displaystyle x^4 - x^3 + (4x^2 + 3x) + 5 =$

$\displaystyle x^4 - x^3 + 4\left(x^2 + \dfrac{3}{4} * x\right) + 5 =$

$\displaystyle x^4 - x^3 + 4\left(x^2 + 2 * \dfrac{3}{8} * x\right) + 5 =$

$\displaystyle x^4 - x^3 + 4\left\{x^2 + 2 * \dfrac{3}{8} * x + \left(\dfrac{3}{8}\right)^2\right\} + 5 - 4\left(\dfrac{3}{8}\right)^2 =$

$\displaystyle x^2(x^2 - x) + 4\left(x + \dfrac{3}{8}\right)^2 + 5 - 4\left(\dfrac{9}{64}\right) =$

$\displaystyle x^2(x^2 - x) + 4\left(x + \dfrac{3}{8}\right)^2 + 5 - \dfrac{9}{16}.$

Thank you