Ivan Shishkin, Oak Grove

Roots of a quadratic equation with a parameter

Reviewed

For which values of the real mm the equation :

mx22(m+1)x+(m+3)=0mx^2-2(m+1)x+(m+3)=0

admits two distinct real solutions ?

Solutions

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Solution by Ancient Tree

0 useful votes

Solution : For m],0[]0,1[m\in ]-\infty,0[\cup ]0,1[.

There is a trap : one is tempted to use the discriminant, which is :
Δ=4(m+1)24m(m+3)=4[(m2+2m+1)(m2+3m)]=4(1m)\Delta=4(m+1)^2-4 m(m+3)=4\left[\left(m^2+2 m+1\right)-\left(m^2+3 m\right)\right]=4(1-m)
so the discriminant is strictly positive if and only if m<1m<1.
However, the discriminant can be only used when the quadratic equation is not degenerate, that is, when m0m\neq 0. For m=0m=0, the equation is linear, with only one solution. So m=0m=0 has to be removed from the admissible mm.

This is a great illustration of why it is important to make sure we fully understand every term in a problem before trying to solve it, since this can change the solution.