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

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The discussion focuses on proving that the roots of the quadratic equation cx^2 + 2ax + b = 0 are not real, given that a, b, and c are positive constants. Participants emphasize the importance of the discriminant, noting that for real roots, it must be greater than zero. They derive inequalities from the discriminants of two other quadratic equations, establishing that b^2 > ac and c^2 > ab. By multiplying these inequalities, they conclude that a^2 - bc < 0, indicating that the roots of cx^2 + 2ax + b = 0 are indeed not real. The conversation highlights the application of quadratic properties and inequalities in determining the nature of roots.
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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!:)
 
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Hello and welcome to MHB, Lytk! :D

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

Please show what you've tried or what your thoughts are on this problem.
 
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?
 
Lytk said:
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:

$$b^2>ac$$

$$c^2>ab$$

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?
 
$b^2>ac$
$c^2>ab$

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 :)
 
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