If [math]x^2-2ax+b=0[/math] ,[math]b\neq 0[/math] and has two unequal real roots then show that [math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\geq 0[/math]

 

I know that since the equation has two unequal real roots then [math]4a^2-4b>0[/math] but then how does this figure in the above inequality??

Some rules that will be of use here:

• If u≥0 and v≥0 then u+v≥0.
  
• If u≥0 and v≥0 then uv≥0.

 

Where do you get these tedious but trivial problems?

This is similar to the polynomial problem you posted earlier. I get the feeling these problems are designed to test not how much you know but rather how good your observational skills are.

 

Hint for this problem: Factorize the LHS as the product of two quadratic expressions in a.

Edited June 4, 201015 yr by shyvera

just find discriminant and [-b-+[math]\sqrt{\Delta}[/math]]/2a

just find discriminant and [-b-+[math]\sqrt{\Delta}[/math]]/2a

 

That won’t help.

 

Further hint:

 

[math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\ =\ \left(a^2-b\right)(\cdots)[/math]

That won’t help.

 

Further hint:

 

[math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\ =\ \left(a^2-b\right)(\cdots)[/math]

 

i think that replacing the root whit x gives the same result as you v written . doesnt it?

Substituting [math]x=a\pm\sqrt{a^2-b}[/math] in [math]x^2-2ax+b=0[/math] yields 0=0, a tautology.

 

That [math]x^2-2ax+b=0[/math] has two distinct real roots is a given. From this, triclino correctly derived the condition [math]4a^2-4b>0[/math]. Stated more succinctly, [math]a^2-b>0[/math].

 

This reduces the problem to

Given [math]a^2-b>0[/math], show that [math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\geq 0[/math].

 

As shyvera already noted, the first step to proving this is to factor the left hand side into something of the form [math](a^2-b)(\cdots)[/math].