- #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##.**