A Polynomial with Two Square Roots

From the Fundamental Theorem of Algebra we know that the polynomial has exactly $4$ complex roots when counted with multiplicity. Note that when $a= 0$ , the polynomial becomes $x^4 - x^2$ , which has exactly $3$ distinct roots:- $0$ , $1$ , and $-1$ . Hence we conclude that $a \neq 0$ . The problem can now be divided into two cases.

Case 1:- One of the roots appears thrice and the other root appears once Let the root that appears thrice be $p$ , and let the other root be $q$ . Then we have $x^4 + (a^2-1)x^2 + a^3 = (x-p)^3(x-q)$ [note that the polynomial is monic]. Comparing coefficients

pic ,

we obtain the following equations.

$3p+q= 0$ $.....(1)$

$p+3q= 0$ $.....(2)$

$3p(p+q)= a^2-1$ $.....(3)$

$p^3q = a^3$ $.....(4)$

Solving equations $1$ and $2$ for $p$ and $q$ gives $(p,q)= (0,0)$ , a contradiction to the fact that $a \neq 0$ . Hence this case yields no solution for $a$ .

Case 2:- both the roots appear twice Let the roots be $p$ and $q$ . Then we must have $x^4 + (a^2-1)x^2 + a^3= (x-p)^2(x-q)^2$ [note that the polynomial is monic]. Then comparing coefficients

(http://s23.postimg.org/mcw12ynln/roots2.png),

we obtain the following equations.

$p+q= 0$ $.....(*1)$

$p^2 + 4pq + q^2 = a^2-1$ $.....(*2)$

$p^2q^2= a^3$ $.....(*3)$

From $(*1)$ , $q= -p$ . Plugging this value in $(*2)$ and $(*3)$ respectively, $-2p^2= a^2 - 1$ and $p^4= a^3$ . Eliminating $p$ from these equations, $a^4 - 4a^3 - 2a^2 + 1= 0$ (note that this will produce an extraneous value for $p^2$ , but the genuine value of $p^2$ will also be included, and since we're interested in the values of $a$ not of $p$ , this equation will give genuine values of $a$ ).

Now we shall prove that the polynomial $f(x) = x^4 - 4x^3 - 2x^2 + 1$ has no repeated roots (this step is necessary or else we might double count some values of $a$ ). Consider $f'(x) = 4x^3 - 12x^2 - 4x$ . For a polynomial to have repeated roots, it must share a common root with its derivative. Now note that $f(x)= \frac{x-1}{4}f'(x) - 4x^2 - x + 1$ . So if $m$ is a common root of $f(x)$ and $f'(x)$ , $m$ must satisfy $-4m^2 - m + 1= 0$ . But again, $f'(m)= \frac{13-4m}{4} (-4m^2 - m + 1) + \frac{m-13}{4}$ . So $m$ must satisfy $\frac{m-13}{4}= 0 \implies m= 13$ . But we see that $f(13) \neq 0$ . Hence the polynomial $f(x)$ has no repeated roots. Thus by Vieta's formula, the sum of the distinct roots if $f(x)$ will be $\frac{4}{1}= 4$ . Hence the sum of all distinct values of $a$ is $4$ .

Let $f_a(x) = x^4 + (a^2-1)x^2 + a^3$ . Then $x = 0$ is a root if and only if $a = 0$ , and in such a case, there are three distinct roots: $f_0(x) = x^4 - x^2 = x^2(x-1)(x+1)$ . Since $f_a(x) = f_a(-x)$ for all $x$ , if $r$ is a nonzero root of $f_a$ , then $-r$ also is a distinct nonzero root. Because $f_a(x) = (x^2)^2 + (a^2-1)x^2 + a^3$ is a quadratic in $x^2$ , we can characterize the complete solution set to $f_a(x) = 0$ as $x \in \pm \sqrt{ \frac{(1-a^2) \pm \sqrt{(a^2-1)^2-4a^3}}{2} },$ where the signs may be chosen independently. Hence for $a \ne 0$ , $f_a$ has four distinct roots unless the expression inside the outer square root is the same regardless of which sign is chosen for the inner square root; that is to say, if the discriminant $D(a)$ is equal to zero: $\begin{aligned} 0 &= D(a) = (a^2-1)^2-4a^3 \\ &= a^4 - 4a^3 - 2a^2 + 1. \end{aligned}$ Hence the sum of all complex $a$ for which $f_a(x)$ has exactly two distinct roots is equal to the sum of the roots of $D(a)$ , which by Vieta's formula, is $\boxed{4}$ .

Let $y = x^{2}$ . Then, the equation becomes $y^{2} + (a^{2} - 1)y + a^{3}$ . Note that if the discriminant is non-zero, the quadratic equation has two distinct complex roots. At least one of these roots is non-zero, and every non-zero complex number has two distinct square roots. So the smallest possible number of distinct complex roots of the original equation is 3, which is not 2. So, the discriminant is zero.

$(a^{2} - 1)^{2} - 4a^{3} = 0$ . The sum of roots of the equation is the coefficient of $a^{3}$ , which is 4.

Let $y = x^2$ . There are either two distinct roots or a double root to $y^2 + (a^2-1)y + a^3 = 0$ [1].

For any complex $y$ , there are exactly $2$ distinct complex solutions to $x^2 = y$ , unless $y = 0$ in which case $x^2 = 0$ has exactly $1$ solution.

If there are two distinct solutions $y_1 \neq y_2$ to [1], then one of them is non-zero, say $y_1$ . Then, $x^2 = y_1 \neq 0$ has 2 distinct complex solutions and $x^2 = y_2$ has at least one solution distinct from the other two, since $y_1 \neq y_2$ . This yields at least $3$ distinct solutions, a contradiction.

Therefore, [1] must have a double root. Hence, the discriminant must be $0$ , i.e. $(a^2-1)^2 - 4a^3 = 0 \leadsto a^4 - 4a^3 - 2a^2 + 1 = 0$ .

It is easy to see that this quartic has four distinct roots. By Vieta's, the sum of these values of $a$ is $4$ .

The expression is a quadratic in $x^2$ , which we could let to be $y$ . Letting the function $f(y) = y^2 + (a^2 - 1)y + a^3$ , and solving for the roots by the Quadratic Formula,

$y = \frac {(1 - a^2) \pm \sqrt{(a^2 - 1)^2 - 4a^3}} {2}$

For $y$ to have exactly one complex value, and hence for $x^2$ to have two different complex values, $(a^2 - 1)^2 - 4a^3 = 0$ . Equivalently, $a^4 - 4a^3 - 2a^2 + 1 = 0$ . By Vieta's formula, the four (complex, in general) solutions of this equation, say $a_1, a_2, a_3,$ and $a_4$ , have a sum equal to $-\frac {-4} {1} = 4$ . Hence, the required answer is $\boxed {4}$ .

Let $t = x^2$ .

Then the equation becomes $t^2 +(a^2-1)t+a^3 = 0$

For the original equation to have two distinct complex roots, the above equation must have repeated roots, i.e, its discriminant is 0.

Thus $(a^2-1)^2 - 4a^3 = 0$ , or $a^4 -4a^3 -2a^2+1 = 0$

This equation has 4 complex roots, the sum of which equals $-(-4/1) = 4$ , which is the negative of second coefficient divided by the first coefficient of the polynomial.

because this equation should have 4 roots,it will have only 2 distinct roots if its a perfect square like (x^2 + k)^2 = 0. Therefore a^2 -1 = 2k a^3= k^2 we get (a^2-1)^2 = 4 (a^3) which is a 4th degree polynomial in a. The sum of roots of such a polynomial is always = - (coefficient of a^3)/(coefficient of a^4) = - (-4)/1 = 4.

Using the quadratic formula, it is clear that the roots of

x ^4 + ( a ^2 - 1) x ^2 + a ^3

are given by:

x ^2 = \frac {-( a ^2 - 1) \pm \sqrt{( a ^2 - 1)^2 - 4(1)( a ^3))}{2(1)}

or

x = \pm \sqrt{ \frac {-( a ^2 - 1) \pm \sqrt{( a ^2 - 1)^2 - 4 a ^3)}{2}}

It is clear that in order to create two distinct roots,

( a ^2 - 1)^2 - 4 a ^3 = 0,

yielding the roots

x = \pm \sqrt{ \frac{-( a ^2-1)}{2}},

must be true.

Thus the sum of the solutions to

( a ^2 - 1)^2 - 4 a ^3 = a ^4 - 4 a ^3 - 2 a ^2 + 1 = 0

gives the sum of all values of a for which the stated polynomial has exactly two distinct complex roots.

Using Vieta's formulas, the sum of these values of a is

frac{- b }{ a } = frac{-(-4)}{1} = 4.

Note that if we let $u=x^2$ the polynomial is reduced to the quadratic $u^2+(a^2-1)u+a^3$ . If $\ u_1, u_2$ are the roots of this polynomial, then $\pm \sqrt{u_1}, \pm \sqrt{u_2}$ will be the roots of our original polynomial in $x$ . Because we want our original polynomial to have exactly two distinct roots, we must have $u_1=u_2 \neq 0$ , so that the two distinct solutions for our original polynomial will be $\pm \sqrt{u_1}$ . In other words the original polynomial has exactly two distinct solutions if and only if the polynomial in $u$ has exactly one nonzero root. A quadratic has exactly one root whenever its discriminant is equal to zero.

$$(a^2-1)^2-4a^3=0$$

$$a^4-4a^3-2a^2+1=0$$

By Vieta's formulas, the sum of the roots of this quartic function is $4$ . This is then the sum of all values of the parameter $a$ for which our original polynomial has exactly two distinct solutions, provided that none of these values of $a$ yield a root of zero in our polynomial in $u$ . To show that this is the case, notice that in order for the polynomial in $u$ to have a root of $0$ , the constant term $a^3$ must be equal to zero, which implies $a=0$ . Note that $a=0$ does not satisfy the quartic polynomial in $a$ , and therefore is not a value for which the polynomial in $u$ has exactly one solution. Therefore, the sum of all values of $a$ is 4.

First, we have to realize that the only way to obtain two distinct roots from this equation is having two double roots. If this form could be factored into a cubic and a linear, say $x+b$ and $(x+a)^3$ , then through multiplying out, we would get either

$a=0$ and $a=\frac{-b}{3}$

or

$a=-3b$ and $9b=b$

which would result in both $a$ and $b$ being equal to $0$ .

That aside, we equate the original equation $x^4+(a^2-1)x^2+a^3$ to the appropriate form for a double quadratic root, namely $(x^2-c)^2$ .

(The only way to prevent $x$ and $x^3$ terms from cropping up is to have the $x$ term in the original quadratic equal $0$ .)

The crucial part begins here. We equate each of the coefficients in the quartic polynomials to each other, resulting in

$a^3=c^2$ (the constant term)

and

$a^2-1=-2c$ (the $x^2$ term)

We solve this system through solving for $c$ in the second equation ( $c=\frac{1-a^2}{2}$ ) and substituting $c$ into the first equation, resulting in

$4a^3=1-2a^2+a^4$

Moving terms, we get the polynomial

$a^4-4a^3-2a^2+1=0$

Finally through Vieta's formulas, we derive the sum of all solutions of this polynomial to be $4$ .

(We do not need to check for extraneous solutions since we did not raise any equation to any power or multiply by a variable.)

If you have x^4 and x^2 as your only x terms in the equation, then if you find any root, it's negative will also be a root. This means that x^4 + (a^2 - 1)x^2 + a^3 better be a perfect square or there will be more than 2 distinct roots.

If it is a perfect square, then [(a^2 - 1)/2]^2 = a^3 a^4 - 4a^3 - 2a^2 + 1 = 0 Sum of solutions to this equation is -(-4/1) = 4.

The sum of the values of a that give 2 distinct roots is 4.

Let $f(x) = x^4 + (a^2-1)x^2 +a^3$ .

$f(x)$ can be expressed as $g(x^2)$ where $g(y) = y^2 + (a^2-1)y + a^3$ .

$g(y)$ is a second-degree polynomial, and thus $g(y) = 0$ has two complex roots (possibly identical). Let's call those roots $y_1$ and $y_2$ . The roots to $f(x) = 0$ are then those $x$ for which $x^2 = y_1$ or $x^2 = y_2$ .

Note that $y = 0$ is not a root to $g(y) = 0$ unless $a = 0$ . $a = 0$ does not contribute to the sum of all $a$ , so we need only consider non-zero roots $y_1$ and $y_2$ .

$x$ such that $x^2 = y_1$ has two solutions. If we call one $x_{11}$ , for the other we have $x_{12} = -x_{11}$ .

Similarly we obtain $x_{21}$ and $x_{22}$ for $x^2 = y_2$ . Since $x_{11}$ and $x_{12}$ are different, and also $x_{21}$ and $x_{22}$ are different, we have either two or four different solutions $x$ , depending on whether $y_1 = y_2$ or not. Since we want two and not four, a necessary and sufficient condition is $y_1 = y_2$ . A quadratic equation $x^2+px+q = 0$ has two identical roots when and only when the discriminant $p^2-4q = 0$ . Here, this means that $(a^2-1)^2 - 4a^3 = 0$ , or $a^4 - 4a^3 -2a^2 + 1 = 0$ .

We do not need to solve this equation, but only determine the sum of all its roots. According to Vieta's formulas, the sum of all roots of a monic polynomial in $x$ of degree $n$ is the negated coefficient of the term $x^{n-1}$ , which is 4. Since all logical steps above are reversible, we haven't introduced any extraneous solutions $a$ , and 4 is our answer.

Firstly, solve $x^4+(a^2-1)x^2+a^3=0$ for $x$ . According to my calculator, there are four roots, which are $x_1=\sqrt(1-a^2+\sqrt(1 - 2 a^2 - 4 a^3 + a^4))/(\sqrt(2))$ ; $x_2=-(\sqrt(1-a^2+\sqrt(1 - 2 a^2 - 4 a^3 + a^4))/(\sqrt(2)))$ ; $x_3=\sqrt(1-a^2-\sqrt(1 - 2 a^2 - 4 a^3 + a^4))/(\sqrt(2))$ ; $x_4=-(\sqrt(1-a^2-\sqrt(1 - 2 a^2 - 4 a^3 + a^4))/(\sqrt(2)))$ Since there are only two distinct roots, therefore $x_1=x_3$ , by estimation, the results are $a_1=-0.476364-0.460077 i, a_2=-0.476364+0.460077 i, a_3=0.51362, a_4=4.43911$ and adding up these four values, the result is $3.9998$ , so the answer is $4$ .

If a grade 4 monic polinomial has 2 distinct complex roots, there are only two possibilities: p(x) = (x-A)^2 (x-B)^2 or p(x)=(x-A)^3 (x-B), for some A, B.

Case 1: p(x) = (x-A)^2 (x-B)^2.

p(x) = x^4 -2(A+B)x^3 + (B^2+4AB+A^2) x^2 -2AB(A+B)x + A^2 B^2.

Since this has to be x^4+(a^2-1)x^2+a^3,

Term of x^3: -2(A+B)=0, with A+B=0. Then, the term of x is also 0, because it has a factor A+B.

Term of x^2: A^2+B^2+4AB = A^2+B^2+2AB+2AB = (A+B)^2+2AB = 2AB, because A+B must be 0. Then, we have the equation 2AB = a^2-1, or AB=(a^2-1)/2.

Term without x: A^2 B^2 = a^3, which means that (AB)^2 = a^3. Taking the value of (AB) from the previous equation, (a^2-1)^2 = 4a^3, which is a 4 grade polinomial, which has 4 complex roots.

Case 2: p(x) = (x-A)^3(x-B):

p(x) = x^4 - (B+3A)x^3 + (3AB+3A^2) x^2 - A^2(3B+A)x+A^3B.

From the x^3 term: B+3A=0.

From the x term: A^2(3B+A)=0. If A=0, then B=0, and there is only one root in p(x). If 3B+A=0, since B+3A=0, the only solution is A=B=0, and again there is only one root in p(x).

So, case 2 is impossible, and case 1 has 4 solution. The result is then, 4 possible values for a.

Finding $x^2$ we get

$x^2= \frac{(1-a^2) \pm \sqrt{(1-a^2)^2 -4a^3}}{2}$

Now, $D\leq 0$ surely implies roots are imaginary.

$(1-a^2)^2-4a^3 \leq 0 \implies a^4-4a^3-2a^2+1 \leq 0$

Sum of values of $a = 4$