Let
be the two roots. The claim is that
![(1)/(x_1)+(1)/(x_2)=(1)/(x_1+x_2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9bpxhi8rfzl7b3cksdkhaqlvg6ser0qxfk.png)
We can rewrite this expression as
![(x_1+x_2)/(x_1x_2)=(1)/(x_1+x_2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gwscxyu64vf2l0r9cxenz34pt7s6ljxql2.png)
Now, recall that if the leading term is 1, then you can think of a quadratic equation as
![x^2-sx+p=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hec45rm6kkrgc22e4elv41wudypjgybweu.png)
i.e. the linear coefficient is the opposite of the sum of the roots, and the constant term is the product of the roots. In other words, we have
![x_1+x_2=m,\quad x_1x_2=1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ewwctej7s4epyyovtg17i3no03mdllttee.png)
Substitute these values in the equation above to have
![(m)/(1)=(1)/(m)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/43f3axj1o0683uf9qzaistc1ue3ppa5v3s.png)
Which leads to
![m^2=1 \iff m=\pm 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hiu8vprwdap2pths4v6823ah3ei4svdz89.png)