+0

solve m, so there is at least one solution in R

0
319
1

mx2 + 2mx = 2x − 1 − m

Oct 12, 2017

#1
+99383
+1

mx2 + 2mx = 2x − 1 − m

$$mx^2 + 2mx = 2x − 1 − m\\ mx^2 + 2mx - 2x +1+ m=0\\ mx^2 + (2m - 2) x+(1+ m)=0\\ \text{This will have at least one solution when }\triangle\ge 0\\ \triangle=b^2-4ac\\ (2m-2)^2-4m(1+m)\ge 0\\ (4m^2-8m+4)-4m-4m^2\ge 0\\ -12m+4\ge 0\\ -12m\ge -4\\ m\le \frac{1}{3}$$

.
Oct 12, 2017