Are the Roots of cx^2 + 2ax +b = 0 Real?

[Lytk]

a, b , c are positive constants and the roots of ax^2 + 2bx+ c
and bx^2 + 2cx +a are all real and unequal(unique).
Show that the roots of cx^2 + 2ax +b = 0 are NOT real.

Help!:)

 

[MarkFL]

Hello and welcome to MHB, Lytk! :D

What does the discriminant of a quadratic function tell us about the nature of its roots?

 

[Euge]

Hi Lytk,

Please show what you've tried or what your thoughts are on this problem.

 

[Lytk]

Hey :)
so the way to prove that the roots are not real, would be if the discriminant is less than 0.
I used the quadratic formula first to find the real roots of the first two equations:

The root of $ax^2 + 2bx +c$ =
$ \frac{-b\pm\sqrt{b^2-ac}}{a}$ ( Real) ${b^2-ac}$ > 0

The root of $bx^2 + 2cx +a$ =

$$\frac{-c\pm\sqrt{c^2-ab}}{c} $$ (Real) ${c^2-ab}$ > 0The roots of $cx^2 + 2ax +b$ =
$$\frac{-a\pm\sqrt{a^2-bc}}{c} $$ IF ${b^2-ac}$ > 0
$b^2$ >ac

IF ${c^2-ab}$ > 0
$c^2$ >ab

Is there some way of proving
${a^2-bc}$ < 0
which would mean the roots are not real?

 

[MarkFL]

Hey :)
so the way to prove that the roots are not real, would be if the discriminant is less than 0.
I used the quadratic formula first to find the real roots of the first two equations:

The root of $ax^2 + 2bx +c$ =
$ \frac{-b\pm\sqrt{b^2-ac}}{a}$ ( Real) ${b^2-ac}$ > 0

The root of $bx^2 + 2cx +a$ =

$$\frac{-c\pm\sqrt{c^2-ab}}{c} $$ (Real) ${c^2-ab}$ > 0The roots of $cx^2 + 2ax +b$ =
$$\frac{-a\pm\sqrt{a^2-bc}}{c} $$ IF ${b^2-ac}$ > 0
$b^2$ >ac

IF ${c^2-ab}$ > 0
$c^2$ >ab

Is there some way of proving
${a^2-bc}$ < 0
which would mean the roots are not real?

You are almost there. :D

You have correctly found that we must have:

\(\displaystyle b^2>ac\)

\(\displaystyle c^2>ab\)

Now, if both sides of two inequalities are positive (and ours are since we are told $a,b,c>0$), then we can multiply the corresponding sides of the inequalities together, to get another valid inequality. So, what do we get when we multiply these 2 inequalities?

 

[Lytk]

$b^2>ac$
$c^2>ab$

I multiply the corresponding sides and I get:
$b^2c^2>a^2cb$
which cancels down to
$bc>a^2$
$0>a^2 -bc$ (discriminant)

$\therefore$ Roots are not realTHANK YOU SO MUCH! This was confusing me for a long time :)