Three () distinct real roots of a quadratic equation

In summary, the equation ##ax^2+bx+c = 0## with three roots ##x_1, x_2, x_3## given to be pairwise distinct, can be shown to have ##a = b = c = 0## by applying the factor theorem to all three roots and using the fact that the polynomial has a degree of 2.
  • #1
brotherbobby
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Homework Statement
You are given the quadratic equation ##ax^2+bx+c=0##. Prove that if ##x_1, x_2\; \text{and}\; x_3## are pairwise distinct real roots of this equation, then ##\boxed{a=b=c=0}##
Relevant Equations
If ##\alpha,\beta## are (real or complex conjugate) roots the quadratic equation ##ax^2+bx+c = 0##, then the sum of the roots ##\alpha+\beta = -\frac{b}{a}## and the product of the roots ##\alpha \beta = \frac{c}{a}##
It is given that ##x_1, x_2\; \text{and}\; x_3## are roots of the equation ##ax^2+bx+c=0##, which are pairwise distinct.

If indeed they are roots, we should have ##ax_1^2+bx_1+c= 0 = ax_2^2+bx_2+c= 0 = ax_3^2+bx_3+c= 0##.

On subtracting the first two, we obtain ##a(x_1^2-x_2^2)+b(x_1-x_2) = 0\Rightarrow (x_1-x_2) \left[a(x_1+x_2)+b\right]=0##. Since ##x_1 \neq x_2##, we must have ##\left[a(x_1+x_2)+b\right] \Rightarrow x_1+x_2 = -\tfrac{b}{a} \enclose{circle}{1}##(not a surprise). Taking the other equations pairwise, we are left with ##x_2+x_3 = -\tfrac{b}{a} \enclose{circle}{2}= x_3+x_1 = -\tfrac{b}{a} \enclose{circle}{3}##.

Taking ##\enclose{circle}{2} - \enclose{circle}{1} \rightarrow x_3-x_1 = 0\; \enclose{circle}{4} ##.

Thus we have ##x_1 = x_2 = x_3## using the other equations pairwise which violate them being pairwise distinct.

I could not proceed further. How do we use this to show what is asked for, viz. ##a = b = c = 0##.
 
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  • #2
You have assumed that ##a\neq 0## (in order to be able to put ##a## in the denominator of ##-\frac{b}{a}## and be able to do all that algebraic manipulations) and you proved that ##x_1=x_2=x_3## which contradicts the given hypothesis that they are pairwise distinct. Therefore it must be ##a=0##.

After that you assume that ##b\neq 0## and similarly you are going to be led to a contradiction. Therefore it must be ##b=0##.

And after that it is fairly easy to prove ##c=0##.
 
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  • #3
You generated a contradiction, assuming your calculations were correct. I think you have two assumptions, one is the roots are district e, and the other is they b/a makes sense. I think the correct conclusion is that a must be zero if the roots are distinct.
 
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  • #4
Another approach is to write it out as a system of equations in matrix form:
$$
\begin{pmatrix}
x_1^2 & x_1 & 1 \\
x_2^2 & x_2 & 1 \\
x_3^2 & x_3 & 1 \\
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c \\
\end{pmatrix} =
\begin{pmatrix}
0 \\
0 \\
0 \\
\end{pmatrix}
$$
Then show that the determinant of that matrix is ##(x_1 - x_2)(x_2 - x_3)(x_1 - x_3)##. And if ##x_1, x_2, x_3## are pairwise distinct, then the matrix is invertible and ##a = b = c = 0##.
 
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  • #5
Delta2 said:
You have assumed that ##a\neq 0## (in order to be able to put ##a## in the denominator of ##-\frac{b}{a}## and be able to do all that algebraic manipulations) and you proved that ##x_1=x_2=x_3## which contradicts the given hypothesis that they are pairwise distinct. Therefore it must be ##a=0##.

After that you assume that ##b\neq 0## and similarly you are going to be led to a contradiction. Therefore it must be ##b=0##.

And after that it is fairly easy to prove ##c=0##.

1. I got the point about ##a = 0##.

2. However, nowhere do we assume that ##b\neq 0## (##b## comes on the numerator). How do we then proceed to show that ##b = 0##?

3. Regarding ##c##. We have the three equations ("product of roots of a quadratic equation") : ##x_1 x_2 = c/a = x_2 x_3 = x_3 x_1##. Again as ##c## appears on the numerator, no assumption is made for ##c \neq 0##.

I don't know how to proceed from here.
 
  • #6
brotherbobby said:
1. I got the point about ##a = 0##.

2. However, nowhere do we assume that ##b\neq 0## (##b## comes on the numerator). How do we then proceed to show that ##b = 0##?

3. Regarding ##c##. We have the three equations ("product of roots of a quadratic equation") : ##x_1 x_2 = c/a = x_2 x_3 = x_3 x_1##. Again as ##c## appears on the numerator, no assumption is made for ##c \neq 0##.

I don't know how to proceed from here.
If you can show that ##a = 0##, then surely you are almost home and dry, as you have a linear equation with three roots.
 
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  • #7
PeroK said:
If you can show that ##a = 0##, then surely you are almost home and dry, as you have a linear equation with three roots.

Thank you. Elementary though it sounds, let me still complete the argument.

Given : We have a quadratic equation ##ax^2+bx+c = 0## with three roots ##x_1, x_2, x_3## given to be pairwise distinct.

We use the equation and the idea of sum of roots ##\left( \alpha+\beta = -b/a \right)## to show that ##\boxed{a = 0}##.

With ##a = 0##, our equation reduces to a linear equation : ##bx + c = 0##. Since ##x_1, x_2## are two of its roots, we must have : ##bx_1+c = bx_2+c = 0 \Rightarrow b(x_1-x_2) = 0 \Rightarrow \boxed{b=0}## since ##x_1\neq x_2##.

If ##b=0##, the linear equation above (in red) reduces to ##\boxed{c=0}##.

Thank you. It's been a learning.
 
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  • #8
do you know the root-factor theorem? that if x1 is a root, then X-x1 is a factor?
 
  • #9
Yes, let me write it out, well-known though it may be. Just to ensure we're both on the same page.

Remainder theorem :
If a polynomial ##P_n(x)## (of order ##n##) is divided by the factor ##(x-a)##, we get the equation : $$P_n(x) = (x-a) Q_{n-1}(x) + R(x)$$ The Remainder theorem states that "the remainder of such a division ##\boldsymbol{R = P_n(a)}##" as one can easily verify by putting ##x=a## in the equation above.

Now we go to what you asked for, the so called Factor theorem.

Factor theorem : If for the above polynomial ##P_n(x)##, ##a## is a root ##a = \alpha##, then the Factor theorem says that "##\boldsymbol{R(\alpha)=0 \Rightarrow P_n(x) = (x-\alpha) Q_{n-1}(x)}##".

Please comment when you can. Also let me know how the Factor theorem helps us in solving the problem statement that I mentioned in this thread in Post # 1.
 
  • #10
You should apply the factor theorem to all three of the roots. It says something about the degree of your polynomial.
 
  • #11
Office_Shredder said:
You should apply the factor theorem to all three of the roots. It says something about the degree of your polynomial.
But how does that help prove ##a=b=c=0##?
 
  • #12
vela said:
But how does that help prove ##a=b=c=0##?

What did you get by applying the theorem?
 
  • #13
Right. I created this thread - let me carry out @Office_Shredder and @mathwonk's strategy of using the factor theorem.

Problem statement : The quadratic equation ##ax^2+bx+c = 0## has three roots ##x_1, x_2, x_3## given to be pairwise distinct. Show that ##a = b = c = 0##
.

Solution (use of factor theorem) : Using the factor theorem, we get ##ax^2+bx+c = (x-x_1)(x-x_2)(x-x_3)##.

Simplifying the expression on the right : \begin{align*}ax^2+bx+c &=(x-x_1)(x-x_2)(x-x_3) \\\Rightarrow ax^2+bx+c &=(x-x_1) (x^2-xx_2-xx_3+x_2x_3)\\ \Rightarrow ax^2+bx+c &=x^3-x^2x_1-x^2x_2+xx_1x_2-x^2x_3+xx_3x_1+xx_2x_3-x_1x_2x_3\\\Rightarrow ax^2+bx+c &=x^3-x^2(x_1+x_2+x_3)+x(x_1x_2+x_2x_3+x_3x_1)-x_1x_2x_3\\ \end{align*}
Taking everything to the left hand side ##x^3 - x^2(a+x_1+x_2+x_3)+x(x_1x_2+x_2x_3+x_3x_1-b)-(x_1x_2x_3+c) = 0##.

I am stuck here. I don't see how does this help me prove that ##a = b = c = 0##.
 
  • #14
You need a constant multiple on the right hand side of your initial equation also. Otherwise the top degree term always has a coefficient of 1.

I don't think you should even put everything on the left hand side. How do you feel about the second to last equation you wrote down (after you fix it). Can you solve all the unknowns (a, b, c and that vl new multiple you are going to add) in terms of the roots?
 
  • #15
It is hard to wrestle with every piece of information all at once. Try using the roots just one at a time, and maybe only show the coefficients are zero one at a time. E.g. first show a=0 by contradiction. I.e. suppose a≠0, and then divide through by it getting X^2 + b'X + c', where b' = b/a and c' = c/a. Now X^2+b'X+c' has the same roots as the original equation (why?). So since we know x1 is a root, we get X^2+b'X+c' = (X-x1)(X-d) for some d. Now what do you get from knowing that x2 and x3 are also roots?
 
  • #16
Office_Shredder said:
How do you feel about the second to last equation you wrote down (after you fix it). Can you solve all the unknowns (a, b, c and that vl new multiple you are going to add) in terms of the roots?
I do not get what you were trying to say. I see my post#13 and can't find which equation you were referring to. Please clarify when you can.
 
  • #17
mathwonk said:
It is hard to wrestle with every piece of information all at once. Try using the roots just one at a time, and maybe only show the coefficients are zero one at a time. E.g. first show a=0 by contradiction. I.e. suppose a≠0, and then divide through by it getting X^2 + b'X + c', where b' = b/a and c' = c/a. Now X^2+b'X+c' has the same roots as the original equation (why?). So since we know x1 is a root, we get X^2+b'X+c' = (X-x1)(X-d) for some d. Now what do you get from knowing that x2 and x3 are also roots?

Doing as you said above, we have the quadratic equation ##ax^2+bx+c=0## reduce to ##x^2+b'x+c' = 0## where ##b'=b/a, c'=c/a##. Let the original equation have ##x_1, x_2## as its roots. Then the reduced equation will also have the same roots for the argument given below in brackets.

[Let the reduced equation have different roots, say ##x'_1,x'_2##. The sun of these roots is ##x'_1+x'_2 = -b' = b/a## and the product of the roots ##x'_1x'_2 = c' = c/a##. We know that a quadratic equation can be written down as ##x^2-(\text{Sum of the roots})x+(\text{Product of the roots})=0## which leads to ##x^2-\frac{b}{a}x+\frac{c}{a} = 0## which on multiplying throughout by ##a## yield the original equation]

Continuing, since we know ##x_1,x_2## are roots, we can write : ##x^2+b'x+c' = (x-x_1)(x-x_2) (x-x_3)= 0## remembering that all roots ##x_1,x_2,x_3## have been given to be pairwise distinct. This leads to the contradiction that ##x_1=x_2=x_3## (see Post#7 above). Hence the assumption that ##a\neq 0## must be false, leading to ##a=0##. This reduces the equation to ##bx+c = 0## with the three given roots and proceeding similarly, we can show that ##b=c=0## (see Post#7 above).

Is this what you asked for?
 
  • #18
You are given that ##x_1##, ##x_2##, and ##x_3## are three roots, but the problem statement doesn't say that these are the only roots (in fact, that would be impossible). So factoring the polynomial as ##(x-x_1)(x-x_2)(x-x_3)## doesn't seem legitimate to me. All you can say is that the factorization is of the form ##p(x)(x-x_1)(x-x_2)(x-x_3)## where ##p(x)## is some (possibly trivial) polynomial.
 
Last edited:
  • #19
brotherbobby said:
I do not get what you were trying to say. I see my post#13 and can't find which equation you were referring to. Please clarify when you can.

You're doing it in post #17 now so don't worry about it.
 
  • #20
jbunniii said:
You are given that ##x_1##, ##x_2##, and ##x_3## are three roots, but the problem statement doesn't say that these are the only roots (in fact, that would be impossible). So factoring the polynomial as ##(x-x_1)(x-x_2)(x-x_3)## doesn't seem legitimate to me. All you can say is that the factorization is of the form ##p(x)(x-x_1)(x-x_2)(x-x_3)## where ##p(x)## is some (possibly trivial) polynomial.
If we go down that road, then it might be better to start with the observation that for a solution ##x_1##: $$ax^2 + bx + c = (x - x_1)(ax - \frac c {x_1})$$Where, without loss of generality, we can assume ##x_1 \ne 0## (as all three roots cannot be zero if they are distinct).
 
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  • #21
@brotherbobby another approach is go straight for the jugular: complete the square and show directly that there are at most two roots, as given by the quadratic formula. This also gives you the conditions for zero, one and two roots. And proves there is no possibility of a third.
 
  • #22
What exactly is allowed here? The fundamental theorem of algebra let's you do this in two lines. But I get the feeling this is meant to build up to that.
 
  • #23
PeroK said:
If we go down that road, then it might be better to start with the observation that for a solution ##x_1##: $$ax^2 + bx + c = (x - x_1)(ax - \frac c {x_1})$$Where, without loss of generality, we can assume ##x_1 \ne 0## (as all three roots cannot be zero if they are distinct).
This seems like the most straightforward approach and perhaps the most suitable in this context (precalculus HW). The factorization theorem requires the division algorithm, which may not be appropriate to assume is available at this level. Whereas your approach only requires the elementary observation that the product of two nonzero numbers is nonzero.
 
  • #24
PeroK said:
@brotherbobby another approach is go straight for the jugular: complete the square and show directly that there are at most two roots, as given by the quadratic formula. This also gives you the conditions for zero, one and two roots. And proves there is no possibility of a third.
But of course there can be a third, which is the point of this problem. But only for a specific set of coefficients for which the quadratic formula won't apply.
 
  • #25
jbunniii said:
But of course there can be a third, which is the point of this problem. But only for a specific set of coefficients for which the quadratic formula won't apply.
The case ##a = 0## needs to be handled separately. ##a = 0 \ \Rightarrow \ b = c = 0## and ##a \ne 0 \ \Rightarrow## at most two solutions.
 
  • #26
Quoting myself: "since we know x1 is a root, we get X^2+b'X+c' = (X-x1)(X-d) for some d. Now what do you get from knowing that x2 and x3 are also roots?"

plug in x2 for X and you get 0 = (x2-x1)(x2-d), so at least one factor on the right must equal zero, so x2 must equal either x1 or d, and since we assumed it does not equal x1, it must equal d. The same argument shows that x3 also equals d, a contradiction to all roots being distinct. so in fact a=0.

now we have bX+c = 0 has 3 distinct roots x1,x2,x3. if we assume b≠0, then solving directly gives X = -c/b, which is only one number and hence cannot equal all 3 distionct roots x1,x2,x3. thus b=0.

now we have the equation to be c = 0, which has no root at all unless c=0, so since there are 3 distinct roots, we must have a=b=c=0.
 
  • #27
Straightforward solution:

Assume ##x_1, x_2, x_3## are distinct solutions to ##p(x) = ax^2 + bx + c = 0##. Without loss of generality, we can assume ##x_1 \ne 0## (as all three roots cannot be zero if they are distinct).
$$ax_1^2 + bx_1 + c = 0 \ \Rightarrow \ ax_1 + \frac c {x_1} = -b$$ $$(x - x_1)(ax - \frac c {x_1}) = ax^2 -(ax_1 + \frac c {x_1})x + c = ax^2 + bx + c$$ Hence: $$p(x) = (x - x_1)(ax - \frac c {x_1})$$ Now:$$p(x_2) = (x_2 - x_1)(ax_2 - \frac c {x_1}) = 0 \ \Rightarrow \ ax_2 = \frac c {x_1}\ \ (\text{as} \ x_1 \ne x_2)$$ and, similarly, $$ax_3 = \frac c {x_1}$$ Hence: $$a(x_2 - x_3) = 0 \ \ \text{and} \ \ a = 0 \ \ (\text{as} \ x_2 \ne x_3) $$ Finally, with ##a = 0## we have: $$bx_1 + c = bx_2 + c = 0 \ \Rightarrow \ b(x_1 - x_2) = 0 \ \Rightarrow \ b = 0 \ \Rightarrow \ c = 0$$
 
  • #28
Alternative solution (I think this is best):

If ##a = 0##, then ##b = c = 0## (as above).

If ##a \ne 0##, then (by completing the square) we have: $$x_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ And we have at most two distinct solutions.

Hence, three distinct solutions implies ##a = b = c = 0##.
 

Related to Three () distinct real roots of a quadratic equation

1. What is a quadratic equation?

A quadratic equation is an algebraic equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is a second-degree polynomial equation and can have up to two distinct real roots.

2. How do you know if a quadratic equation has three distinct real roots?

A quadratic equation has three distinct real roots if its discriminant (b^2 - 4ac) is greater than zero. This means that the equation has two solutions that are not equal and can be represented by three distinct points on the x-axis.

3. Can a quadratic equation have more than three distinct real roots?

No, a quadratic equation can only have a maximum of two distinct real roots. This is because the fundamental theorem of algebra states that a polynomial equation of degree n can have at most n distinct roots.

4. What is the significance of having three distinct real roots in a quadratic equation?

Having three distinct real roots in a quadratic equation means that the equation has two distinct solutions that are not equal. This can be useful in solving real-world problems that involve finding multiple solutions.

5. How do you find the three distinct real roots of a quadratic equation?

To find the three distinct real roots of a quadratic equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Simply plug in the values of a, b, and c into the formula and solve for x. The resulting values of x will be the three distinct real roots of the equation.

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