I am working on some math problems from math books, and I have seen this in one math problem.

$x_1^2+x_2^2= (x_1+x_2)^2 - 2x_1x_2$

I simply don't understand this. Can you please explain this to me.

The whole math problem is: You have this equation

$x^2 - 3x + m^2 -1 = 0$

For what parameter m is:

$x_1^2 + x_2^2 > 3$

  • $\begingroup$ are $x_1$ and $x_2$ the two roots to the quadratic equation $x^2-3x+m^2-1=0$ ? $\endgroup$ – bonsoon Jan 3 '13 at 22:37
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    $\begingroup$ Assuming that $x_1$ and $x_2$ are the two roots, then the reason why your initial equation was written in the form it is is because the values $x_1+x_2$ and $x_1x_2$ can be read off directly from the coefficients of the quadratic: to see this, write the quadratic equation as $(X-x_1)(X-x_2) = 0$ and multiply out the left hand side, then compare with your original form $X^2-3X+m^2-1=0$. $\endgroup$ – Steven Stadnicki Jan 3 '13 at 22:40

I think what's going on is that you want to find all values of the constant $m$ so that, if $x_1$ and $x_2$ are the two (not necessarily distinct) solutions to the equation $x^2 - 3x + m^2 -1 = 0,$ then $x_{1}^{2} + x_{2}^{2} > 3.$

Recall Vieta's formulas for the sum and product of the solutions of the quadratic equation $ax^2 + bx + c = 0.$ They say that the sum of the solutions equals $-\frac{b}{a}$ and the product of the solutions equals $\frac{c}{a}.$

Thus, in the case of your quadratic equation, we have $x_{1} + x_{2} = 3$ and $x_{1}x_{2}=m^2 - 1.$ Since $x_{1}^{2} + x_{2}^{2} = \left(x_1 + x_2\right)^2 - 2x_{1}x_{2},$ it follows that $x_{1}^{2} + x_{2}^{2} = 3^2 - 2(m^2 - 1) = 11 - 2m^2.$ Therefore, $x_{1}^{2} + x_{2}^{2} > 3$ becomes $11 - 2m^2 > 3,$ or $m^2 < 4,$ or $-2 < m < 2.$

  • $\begingroup$ Probably a stupid question from my side, but how did you get that x1^2+x2^2=(x1+x2)^2−2x1x2 $\endgroup$ – user55101 Jan 3 '13 at 23:07
  • $\begingroup$ its called a perfect square or square of a sum. Its a very common factorization. $\endgroup$ – Jorge Fernández Hidalgo Jan 3 '13 at 23:13
  • $\begingroup$ Thanks. I didnt recognized it without parenthesis. Thanks. $\endgroup$ – user55101 Jan 3 '13 at 23:19
  • $\begingroup$ More generally, there is a theorem that any perfectly symmetrical polynomial function of $n$ variables (in this case, the function $f(x_1,x_2) = x_1^2+x_2^2$) can be written as a polynomial in certain elementary symmetric polynomials that correspond (in this case) to the coefficients of your quadratic. See en.wikipedia.org/wiki/… for more details. $\endgroup$ – Steven Stadnicki Jan 4 '13 at 0:15

from Viete formulas $$x_1+x_2=3$$ $$x_1\cdot x_2=m^2-1$$ $$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=3^2-2(m^2-1)>3$$ $$2m^2-2<6,m^2-4<0$$ $$(m-2)(m+2)<0$$ $$m\in(-2,2)$$


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