# Find Real Solutions to x4-2x3+kx2+px+36 = 0

• MHB
• DragonMaths
In summary: As for the constants, they are $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.In summary, the solutions to x4-2x3+kx2+px+36 = 0 are x=-3i and x=3i.
DragonMaths
One of the solutions to
x4-2x3+kx2+px+36 = 0 is x = 3
i

Prove that this polynomial has no real solutions (roots) and find the real values of k and p.
-------------------------------------------------------------------------------------------------------------------

So far the only progress I seem to have made was possibly finding the value of p.

Since x - 3i is a solution , we can say that x + 3i is also a solution.

I then proceeded to replacing x with 3i on one side, and then setting the solution to itself but with x as -3i , as follows :

x4-2x3+kx2+px+36 = x4-2x3+kx2+px+36
(3i)4-2(3i)3+k(3i)2+p(3i)+36 = (-3i)4-2(-3i)3+k(-3i)2+p(-3i)+36
81i4-54i3+9ki+3pi+36 = 81i4+54i3+9ki-3pi+36
0 = 108i3-6pi
0=-108i-6pi
0=-(108-6p)i
0=108-6p
6p=108
p=18

This is as much progress as I have made , and I cannot find anything that can help me
Keep in mind this is asked pre-calculus on on 11th grade Advanced Math course with university grade math problems and so on.Any help would be grately appreciated as I have no idea how to continue at this moment.

Last edited:
Are you sure about the sign of $p$?

Since $3i$ and $-3i$ have to satisfy the equation separately, substitution gives you a system of two independent linear equations in the two unknowns $p$ and $k$. As a starter, from this you can solve both $p$ and $k$.

DragonMaths said:
One of the solutions to
x4-2x3+kx2+px+36 = 0 is x = 3
i

Prove that this polynomial has no real solutions (roots) and find the values of k and p.
-------------------------------------------------------------------------------------------------------------------

So far the only progress I seem to have made was possibly finding the value of p.

Since x - 3i is a solution , we can say that x + 3i is also a solution.

I then proceeded to replacing x with 3i on one side, and then setting the solution to itself but with x as -3i , as follows :

x4-2x3+kx2+px+36 = x4-2x3+kx2+px+36
(3i)4-2(3i)3+k(3i)2+p(3i)+36 = (-3i)4-2(-3i)3+k(-3i)2+p(-3i)+36
81i4-54i3+9ki+3pi+36 = 81i4+54i3+9ki-3pi+36
0 = 108i3-6pi
0=-108i-6pi
0=-(108-6p)i
0=108-6p
6p=108
p=18

This is as much progress as I have made , and I cannot find anything that can help me
Keep in mind this is asked pre-calculus on on 11th grade Advanced Math course with university grade math problems and so on.Any help would be grately appreciated as I have no idea how to continue at this moment.

Hi DragonMaths, welcome to MHB! (Wave)

Indeed, if $x=3i$ is a solution then $x=-3i$ is also a solution assuming that all coefficients are real numbers.
It means that we can factorize the polynomial.
And we can pick as a known factor $(x-3i)(x+3i)=(x^2+9)$.

So we can write the polynomial as:
$$(x^2+9)(x^2+ax+b) = x^4+ax^3+(b+9)x^2+9ax+9b$$

It follows that $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.
Can we find the constants from that?
Does the remaining $x^2+ax+b=0$ have any real roots? (Wondering)

Note that the (as usual correct!) post #3 gives you a way to obtain the quotient of the polynomial $p(x) = x^4 - 2x^3 + kx^2 + px + 36$ upon division by the known factor $m(x) = x^2 + 9$. Namely, it sets you up for solving a system of equations for the coefficients of $q$.

Another way to obtain $q$ is by long division of $p$ by $m$. I don't know if this is part of your pre-calculus course, but here it usually is, and it is a useful technique to know.

I like Serena said:
Hi DragonMaths, welcome to MHB! (Wave)

Indeed, if $x=3i$ is a solution then $x=-3i$ is also a solution assuming that all coefficients are real numbers.
It means that we can factorize the polynomial.
And we can pick as a known factor $(x-3i)(x+3i)=(x^2+9)$.

So we can write the polynomial as:
$$(x^2+9)(x^2+ax+b) = x^4+ax^3+(b+9)x^2+9ax+9b$$

It follows that $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.
Can we find the constants from that?
Does the remaining $x^2+ax+b=0$ have any real roots? (Wondering)

Firstly,thanks for welcoming me :) .Secondly , the question states that we have to prove all the factors are complex,which answers your question that the remaining $x^2+ax+b=0$ would not have any real roots as it is a factor.Thanks you for your contribution.

DragonMaths said:
Firstly,thanks for welcoming me :) .Secondly , the question states that we have to prove all the factors are complex,which answers your question that the remaining $x^2+ax+b=0$ would not have any real roots as it is a factor.Thanks you for your contribution.

Indeed, so we still need to find those values for a and b, and verify that the discriminant, which is $D=a^2-4b$ in this case, is negative, yielding complex solutions. (Nerd)

Krylov said:
Note that the (as usual correct!) post #3 gives you a way to obtain the quotient of the polynomial $p(x) = x^4 - 2x^3 + kx^2 + px + 36$ upon division by the known factor $m(x) = x^2 + 9$. Namely, it sets you up for solving a system of equations for the coefficients of $q$.

Another way to obtain $q$ is by long division of $p$ by $m$. I don't know if this is part of your pre-calculus course, but here it usually is, and it is a useful technique to know.

So I have tried using long division and synthetic division,but since we have to either divide using a complex number (which is not too much of a problem) or we use the known factor of $x^2 + 9$ ,but the unknown coefficients are still both present in the new polynomial created, namely being $x^2 -2x + k - 9$ with a remainder of $(p + 18)x - 9k +117$ which should be equal to 0 , as there should be no remainder when dividing by a factor.But as both $p$ and $k$ are present in the remainder , I'm stumped.

DragonMaths said:
So I have tried using long division and synthetic division,but since we have to either divide using a complex number (which is not too much of a problem) or we use the known factor of $x^2 + 9$ ,but the unknown coefficients are still both present in the new polynomial created, namely being $x^2 -2x + k - 9$ with a remainder of $(p + 18)x - 9k +117$ which should be equal to 0 , as there should be no remainder when dividing by a factor.But as both $p$ and $k$ are present in the remainder , I'm stumped.

Indeed. The remainder must be zero.
So that puts restrictions on p and k.
It means for starters that $p=-18$ (with the correct sign now, as Krylov pointed out).
Otherwise $3i$ and $-3i$ wouldn't be solutions.

Btw, is it given that k and p are real numbers?
As it is, that is not given in the problem statement.
And otherwise we cannot assume that $x=-3i$ is a solution.

I like Serena said:
Hi DragonMaths, welcome to MHB! (Wave)

Indeed, if $x=3i$ is a solution then $x=-3i$ is also a solution assuming that all coefficients are real numbers.
It means that we can factorize the polynomial.
And we can pick as a known factor $(x-3i)(x+3i)=(x^2+9)$.

So we can write the polynomial as:
$$(x^2+9)(x^2+ax+b) = x^4+ax^3+(b+9)x^2+9ax+9b$$

It follows that $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.
Can we find the constants from that?
Does the remaining $x^2+ax+b=0$ have any real roots? (Wondering)

Also , this is a great way to solve it and I did not even think that would so simple.It seems like a valid way to get the coefficients,and yet I've never seen someone do it in that way?

- - - Updated - - -

I like Serena said:
Indeed. The remainder must be zero.
So that puts restrictions on p and k.
It means for starters that $p=-18$ (with the correct sign now, as Krylov pointed out).
Otherwise $3i$ and $-3i$ wouldn't be solutions.

Btw, is it given that k and p are real numbers?
As it is, that is not given in the problem statement.
And otherwise we cannot assume that $x=-3i$ is a solution.

Just updated the original post thank you,the question does state that it is indeed a real number.

DragonMaths said:
Also , this is a great way to solve it and I did not even think that would so simple.It seems like a valid way to get the coefficients, and yet I've never seen someone do it in that way?

It's just one of the many tricks of the trade.
A textbook may focus on explaining long division, and leave this approach out.
It's certainly easier to explain, and I've always preferred the simplest, most intuitive approach. ;)

So after calculating and factorising a little it seems that $x^4 - 2x^3 + 13x^2 - 18x - 36 = 0$ is the solution to the problem , with $k = 13$ and $p = - 18$ for the coefficients , and the roots of the polynomial being equal to $(x + 3i)(x - 3i)(x - 1 + \sqrt{3}i)(x - 1 - \sqrt{3}i)$

Yep. That's what I found as well. (Nod)

Alternatively (and assuming $p$ and $k$ are real),

Substituting $3i$ and $-3i$ for $x$ in $f(x)=x^4-2x^3+kx^2+px+36$ we arrive at

$$81+54i-9k+3ip+36=0$$

and

$$81-54i-9k-3ip+36=0$$

$$162-18k+72=0\implies k=13$$

It can then be found that $p=-18.$

Now, if the other two roots were purely imaginary, $f(x)$ would have no odd powers of $x$ so we have

$$f(x)=(x^2+9)(x-(a+bi))(x-(a-bi))=(x^2+9)(x^2-(a+bi)x-(a-bi)x+(a+bi)(a-bi))=(x^2+9)(x^2-2ax+a^2+b^2)$$

Now, equating coefficients with $x^4-2x^3+13x^2-18x+36$, we have

$$9(a^2+b^2)=36\implies a^2+b^2=4$$

$$-2a=-2\implies a=1\Rightarrow b=\pm\sqrt3$$

The roots of $x^2-2x+4=0$ are $1\pm\sqrt3i$, as required.

## 1. What is the purpose of finding real solutions to this equation?

The purpose of finding real solutions to this equation is to determine the values of x that make the equation true, also known as the roots or solutions of the equation. These solutions can provide valuable information about the behavior of the equation and can be used to solve real-world problems.

## 2. How many solutions does this equation have?

The equation has 4th degree, or quartic, polynomial, therefore it can have up to four solutions. However, not all of these solutions may be real numbers.

## 3. What does the value of k and p represent in this equation?

The values of k and p represent the coefficients of the quadratic and linear terms, respectively. These coefficients affect the shape, location, and orientation of the graph of the equation.

## 4. How can I find the solutions to this equation?

There are several methods for finding solutions to a quartic equation, such as factoring, completing the square, or using the quadratic formula. You can also use a graphing calculator or computer software to approximate the solutions.

## 5. Can this equation have imaginary solutions?

Yes, it is possible for this equation to have imaginary solutions, especially if the coefficients k and p are complex numbers. Imaginary solutions can be represented by the square root of a negative number, and they involve the use of complex numbers in mathematics.

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