# Re: [isabelle] nonnegative quadratic polynomial

```Dear Omar,

Here you are,

lemma
fixes a b c x :: real
assumes  "a > 0"
and "âx. a*x^2 + b*x + câ0 "
shows "discrim a b c â0"
proof -
have "âx. a*x^2 + b*x + c<0 " when "discrim a b c > 0" "a>0"
proof -
let ?P="Îx. a*x^2 + b*x +c"
have "?P (- b / (2*a)) =  (4*a*c-b^2)/(4*a)" using âa>0â
by algebra
also have "... <0"
using that unfolding discrim_def by (simp add: divide_neg_pos)
finally show ?thesis by blast
qed
then show ?thesis using assms
using not_less by blast
qed

Best,
Wenda

On 2016-07-19 15:10, Omar Jasim wrote:
```
```Dear all,

```
equation the discriminant will be nonpositive. I found a proposition in a
```text book bellow :

```
but for this one, the equality hold iff x=-b/2*a and I don't want to put this as an assumption as I need this lemma in another work. the lemma I'm
```trying is:

lemma
fixes a b c x :: real
assumes  "a > 0"
and "âx. a*x^2 + b*x + câ0 "
shows " discrim a b c â0"

I proved that for strictly positive equation as bellow:

lemma
fixes a b c x :: real
assumes  "a > 0"
and "âx. (a*(x)^2 + b*x + c) >0 "
shows " discrim a b c â0"
using assms by (metis discriminant_pos_ex less_le not_less)

```
but for nonnegative equation (the first lemma) I couldn't and way. please could any one help me in this because I spend alot of time trying to prove
```it but unfortunately I failed.

Omar
```
```
--
Wenda Li
PhD Candidate
Computer Laboratory
University of Cambridge

```

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