Hi, it is not clear that your 4th order polynomial can be written as the square of a 2nd order one. What do you want to do ? Find the roots of the 4th oder polynomial ?

Not sure, but I believe Tarek Khalifa want to analyze a quartic equation, i.e. discuss the solutions of it...

For example, as you can check, in a case that such form is possible, then coefficient a shall satisfy a condition: a < 0...

Just to be clear, Tarek Khalifa wrote:

f(X)=-(m*X^2+n*X+y)^2

So, we have - sign in front of left bracket...

A-5. A necessary and sufficient condition

Answer 5, posted on September 17, 2019.

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In the more general case of polynomials with complex coefficients

and complex roots, a polynomial p of degree 4 can be expressed

as negative the square of a polynomial q of degree 2

if and only if either p has 2 roots repeated twice

or one root repeated 4 four times.

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Thanks to properties of the field C of complex numbers

to be an algebraic closed field, the polynomial q

can be expressed as

---

q(x) = m(x – r)(x – s),

---

where m, r, and s are complex numbers, and m is not zero.

If p is negative the square of q, then

---

p(x) = -q(x)^2 = -m^2(x – r)^2(x – s)^2,

---

so that if r is not equal to s,

then r and s are roots of p repeated twice,

and if r = s, then r = s is a root repeated four times.

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Conversely, if p is polynomial of degree 4

with two roots r and s repeated twice,

then p has the form:

---

p(x) = a(x – r)^2(x – s)^2

---

where a is a nonzero complex number.

Then there exists a complex number m such that -a = m^2,

and p is negative the square of the polynomial q defined as

q(x) = m(x – r)(x – s).

If r = s, then r = s is a root of p repeated 4 times.

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With best regards, Jean-Claude

Richard Epenoy

Thanks for your help. Really i want to prove that f(X) is negative definite and find the condition which will make f(X) is negative definite.

Anton Vrdoljak

Thanks for your help. Yes you are right, i want to find the condition which makes f(X) is negative definite.

Richard Epenoy Mercedes Orús-Lacort

We may choose an appropriate form for f(X) other than the form proposed in the question to obtain the condition which verifies the negative definiteness of f(X).

Jean-Claude Evard

Thanks for your detailed answer.

Hi, if you want to satisfy f(x)

Richard Epenoy

a,b,c,d and e are functions of an other variable parameters.

Mercedes Orús-Lacort

Thank you for this useful and detailed answer.

I will follow the methods mentioned in your comment.

A-15. Additional conditions for a quartic polynomial

to take negative real values at every real point

Answer 15, posted on September 18, 2019.

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I thank Tarek for the answer 9.

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In addition to the conditions posted by Mercedes

in answer 12, here are some more conditions

for a quartic polynomial to take negative real values

at every real point.

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Let us start with the complex case

and then let us move to the case of real values:

Let a, b, c, d, and e be complex numbers

such a is not equal to zero.

For every complex number z, let

p(z) = az^4 + bz^3 + cz^2 + dz + e.

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1. If for every real number x the value p(x) is negative,

this implies that for every real number x,

the value p(x) is a real number.

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2. We have p(x) is a real number for every

real number x if and only if the coefficients

a, b, c, d, and e are all real numbers.

So, let us assume that these 5 coefficient are real.

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3. We have that for every real number x

p(x) is negative if and only if all the graph of p

is below the x-axis.

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4. In particular, p(x) must go to negative infinity

when x goes to infinity

and also when x goes to negative infinity.

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5. We have this if and only if the coefficient a is negative.

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6. As said by Mercedes in answer 12, the conditions

for the graph of p to never cross the x-axis,

that is, the condition for p to not have any real root

are given the following Web page of the Encyclopedia Wikipedia:

https://en.wikipedia.org/wiki/Quartic_function

The conditions are given in the following part of the above page:

https://en.wikipedia.org/wiki/Quartic_function#Nature_of_the_roots

We see on this part of the page that

the polynomial p does not have any real root

if and only if DELTA > 0 and P > 0 and D > 0,

where DELTA, P and D are defined on this part of the page

above these 3 conditions.

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7. A first possible first test before checking the above 3 conditions

is to check whether p(x) is negative at the real part

of the two roots of the second derivative of p.

The first derivative is

p’(x) = 4ax^3 + 3bx^2 + 2cx + d.

The second derivative is

p’’(x) = 12ax^2 + 6bx + 2c.

The common real part of the two roots of p’’ is

w = (-6b)/(24a) = -b/(4a).

If p(w) < 0. then we cannot conclude here,

and we have to apply Part 6 above.

If p(w) = 0 or if p(w) > 0, then we have found

a point, namely the point w, where p(w) is not negative.

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With best regards, Jean-Claude

Dear Tarek Khalifa

To get your request you need the following conditions to be satisfied:

(3b²-8ac)/(bd-16ae) = (bc-6ad)/(cd-6be) = (bd-16ae)/(3d²-8ce)

with all coefficients a, c, and e are negative values.

The proof ( Using the resultant techniques)

Example:

f(x) = -(x²+3x+1)² = -x⁴ - 6x³ -11x² - 6x -1

a = -1, b = -6, c = -11, d = -6, e = -1

Check the conditions:

(3(-6)²-8(-1)(-11)/(-6)(-6)-16(-1)(-1))= 1

((-6)(-11)-6(-1)(-6))/(-11)(-6)-6(-6)(-1))= 1

((-6)(-6)-16(-1)(-1))/(3(-6)²-8(-11)(-1))= 1

Best regards