not sure about my answer inInequality

Feb 2016
24
0
USA
The problem is:
a,b,c are edges of triangle,
prove that the equation: ax^2+2(b+c)x+a=0 . has two distinct real root
​and my answer is:
a,b,c are edges therefore they are absolute values.
so
a>0
b>0
c>0

also
Δ=4(b+c)^2-4a^2
and I need to prove that Δ>0
so
4(b+c)^2-4a^2>0
which led me for:
(b+c)^2>a^2
b+c>a
now if so far i'm correct, than all I've to do is say that because of triangle inequality "b+c>a" is correct.
And I done with the proving.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
The problem is: a,b,c are edges of triangle,
prove that the equation: ax^2+2(b+c)x+a=0 . has two distinct real root.
You really need to work on terminology.
Triangles have sides not edges. You also mean that $a,~b,~\&~c$ are the lengths of the sides. That is the only that they are positive real numbers.

Moreover, you start by considering $ax^2+2(b+c)x+a=0$. If its discriminate, $\Delta\ne 0$ then that quadratic has two real roots.

What happens if $\Delta = 0~?$ that is proof by contradiction.
 
Feb 2016
24
0
USA
You really need to work on terminology.
Triangles have sides not edges. You also mean that $a,~b,~\&~c$ are the lengths of the sides. That is the only that they are positive real numbers.

Moreover, you start by considering $ax^2+2(b+c)x+a=0$. If its discriminate, $\Delta\ne 0$ then that quadratic has two real roots.

What happens if $\Delta = 0~?$ that is proof by contradiction.
First of all, thank you for the quick replay. It is my mistake, I didn't mention that English was not my first language, so basically I do understand the terminology but not how to translate it properly :p
I might don't translate it accurately but in this problem, I've asked to proof that the quadratic has two real roots. Therefore, the only way I know is to proof that is by:
  1. a \(\displaystyle \ne\) 0 , which I know because it is part of the triangle
  2. \(\displaystyle \Delta \)> 0 That I've proven as seen above. My question basically is, if my proof for that condition are correct.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
  1. a \(\displaystyle \ne\) 0 , which I know because it is part of the triangle
  2. \(\displaystyle \Delta \)> 0 That I've proven as seen above. My question basically is, if my proof for that condition are correct.
Δ=4(b+c)^2-4a^2
and I need to prove that Δ>0
so
4(b+c)^2-4a^2>0
which led me for:
(b+c)^2>a^2
b+c>a
now if so far i'm correct, than all I've to do is say that because of triangle inequality "b+c>a" is correct.
Frankly, I would not give you much credit for that so-called proof.
Agreed: you do want to prove that $\Delta>0$ but it cannot be done by assuming that it is already true.
 
Feb 2016
24
0
USA
Well I got stuck (Crying), I can't find a solution to proof that \(\displaystyle \Delta \neq 0\) while \(\displaystyle a,b,c>0\)
Any help here please.
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$\Delta = [2(b+c)]^2 - 4a^2$

$\Delta = 4(b+c)^2 - 4a^2$

$\Delta = 4[(b+c)^2 - a^2]$

$\Delta = 4(b+c+a)(b+c-a)$

$4 > 0$, $(b+c+a)>0$, and $(b+c)>a \implies (b+c-a)>0$

since all three factors are positive, $\Delta > 0 \implies ax^2+2(b+c)x+a=0$ has two distinct real roots.
 
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Feb 2016
24
0
USA
\(\displaystyle (b+c)>a \implies (b+c-a)>0$\)
But if the parameters are:
\(\displaystyle a=3\)
\(\displaystyle b=2\)
\(\displaystyle c=1\)

then \(\displaystyle b+c-a=0\)
that led me into \(\displaystyle \Delta = 0\) which I don't want.
Or am I missing something in your answer?