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Find the number of real roots

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,706
Hi MHB,

This is the second headache problem that I wish to get some insight from MHB today...

Problem:
It is known that the equation \(\displaystyle ax^3+bx^2+cx+d=0\) has three distinct real roots. How many real roots does the following equation have?

\(\displaystyle 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\)

Attempt:

I observed that if we let \(\displaystyle f(x)=ax^3+bx^2+cx+d\), then the given second equation can be rewritten as

\(\displaystyle 6f(x)f''(x)=(f'(x))^2\)

That is, the LHS of the equation has a total of 4 roots, and the RHS of the equation has a total of 2 repeated roots and the range of it is $[0, \infty)$. But I know I need to figure out how many times the curve of the LHS function cuts the curve of the RHS function, but not finding out their roots.

I don't see even a single way to proceed from there and I feel so supremely stupid now.

Can anyone please help me?

Thanks.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,712
Problem:
It is known that the equation \(\displaystyle ax^3+bx^2+cx+d=0\) has three distinct real roots. How many real roots does the following equation have?

\(\displaystyle 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\)

Attempt:

I observed that if we let \(\displaystyle f(x)=ax^3+bx^2+cx+d\), then the given second equation can be rewritten as

\(\displaystyle 6f(x)f''(x)=(f'(x))^2\) That 6 should be a 2.

That is, the LHS of the equation has a total of 4 roots, and the RHS of the equation has a total of 2 repeated roots and the range of it is $[0, \infty)$. But I know I need to figure out how many times the curve of the LHS function cuts the curve of the RHS function, but not finding out their roots.
We are not told how many real roots the quartic equation should have, so let's start with a simple numerical experiment. The simplest cubic polynomial with three distinct real roots is $f(x) = x^3 - x$ (with roots $0,\,\pm1$). Then the equation $2f(x)f''(x)=(f'(x))^2$ becomes \(\displaystyle 2(x^3-x)(6x) = (3x^2-1)^2\). That simplifies to $3x^4 - 6x^2 - 1 = 0$, a quadratic with real solutions $x^2 = \frac16(6\pm4\sqrt3)$. one of those solutions for $x^2$ is positive, the other one is negative. So when we take square roots, we see that the quartic equation has two real roots. Thus it looks as though the answer to our problem should be that the equation \(\displaystyle 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\) has two real roots.

When doing that calculation, I noticed that if I differentiate the quartic polynomial $3x^4 - 6x^2 - 1$ then I get $12(x^3-x)$, which is exactly 12 times the polynomial that I started with. Interesting!

Now let $f(x)$ be a cubic polynomial with three distinct real roots. Since those roots are not repeated, the derivative $f'(x)$ must be nonzero at each of them. Let $g(x) = 2f(x)f''(x) - \bigl(f'(x)\bigr)^2$. Then $g'(x) = 2f(x)f'''(x) + 2f'(x)f''(x) - 2f'(x)f''(x) = 2f(x)f'''(x).$ But $f'''(x)$ is a nonzero constant. Therefore $g'(x)=0$ at the points where $f(x)=0$; and if $f(x) = 0$ then $g(x) = -\bigl(f'(x)\bigr)^2 < 0.$

What that tells us is that the quartic polynomial $g(x)$ has three turning points (which must be a local min., then a local max., then a local min.), and at each of them $g(x)$ is negative. On the other hand, $g(x)\to\infty$ as $x\to\pm\infty$. It follows that the graph of $g$ crosses the $x$-axis exactly twice, and so the equation $g(x)=0$ has two real roots.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,706
Thank you so so much for the reply, Opalg!

We are not told how many real roots the quartic equation should have, so let's start with a simple numerical experiment.
Hmm...I don't quite get this, Opalg. The problem told us $f(x)$ has three distinct real roots, let's say it takes the form $f(x)=a(x-p)(x-q)(x-r)$, then when we multiply it with another linear function, says, $(x-k)$, this yields a quartic expression which when we set it equals zero, a quartic function of $g(x)=a(x-p)(x-q)(x-r)(x-k)$ and this function surely has 4 distinct real roots...Ah, I see where it went wrong already! We are not entirely sure if the value of $k$ matches any one of the three distinct real roots of the function of f! I'm sorry, Opalg! And I'll be more careful next time, try my best not to assume something which isn't correct thus leads to a bunch of useless work. (I mean to say, I even drew some graphs out on my own before asking for help at MHB based on my silly assumption.)

The simplest cubic polynomial with three distinct real roots is $f(x) = x^3 - x$ (with roots $0,\,\pm1$). Then the equation $2f(x)f''(x)=(f'(x))^2$ becomes \(\displaystyle 2(x^3-x)(6x) = (3x^2-1)^2\). That simplifies to $3x^4 - 6x^2 - 1 = 0$, a quadratic with real solutions $x^2 = \frac16(6\pm4\sqrt3)$. one of those solutions for $x^2$ is positive, the other one is negative. So when we take square roots, we see that the quartic equation has two real roots. Thus it looks as though the answer to our problem should be that the equation \(\displaystyle 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\) has two real roots.

When doing that calculation, I noticed that if I differentiate the quartic polynomial $3x^4 - 6x^2 - 1$ then I get $12(x^3-x)$, which is exactly 12 times the polynomial that I started with. Interesting!
WOW...I like your way of approaching the problem by trying some simple cubic function out in the first place...and I chide myself why I thought of doing the same for another problem (another headache problem of the same type) but not for this one, though I haven't cracked that problem yet!

You have made my understanding to the problem to 100%!

Now let $f(x)$ be a cubic polynomial with three distinct real roots. Since those roots are not repeated, the derivative $f'(x)$ must be nonzero at each of them. Let $g(x) = 2f(x)f''(x) - \bigl(f'(x)\bigr)^2$. Then $g'(x) = 2f(x)f'''(x) + 2f'(x)f''(x) - 2f'(x)f''(x) = 2f(x)f'''(x).$ But $f'''(x)$ is a nonzero constant. Therefore $g'(x)=0$ at the points where $f(x)=0$; and if $f(x) = 0$ then $g(x) = -\bigl(f'(x)\bigr)^2 < 0.$

What that tells us is that the quartic polynomial $g(x)$ has three turning points (which must be a local min., then a local max., then a local min.), and at each of them $g(x)$ is negative. On the other hand, $g(x)\to\infty$ as $x\to\pm\infty$. It follows that the graph of $g$ crosses the $x$-axis exactly twice, and so the equation $g(x)=0$ has two real roots.
I must admit, even if I tried out myself by using the example of some simple cubic equation, I know I won't think of differentiating the equation $g(x)=2f(x)f''(x)-(f'(x))^2$ again, just to study the behavior of the curve of g(x) to draw the conclusion on how many times the function of $g(x)$ cuts the x-axis!

I've learned a great deal from you ever since I joined MHB, and you're my hero! I truly like you, as a mentor!
 
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