Question Number 13359 by ajfour last updated on 19/May/17
$${For}\:{two}\:{real}\:{and}\:{distinct} \\ $$$${solutions}\:{to}\:: \\ $$$${y}=\mathrm{3} \\ $$$${y}={ax}^{\mathrm{2}} +{b} \\ $$$${As}\:{can}\:{be}\:{seen}\:{from}\:{graph}\:{in} \\ $$$${comment}\:{below}, \\ $$$${if}\:{a}>\mathrm{0},\:{b}<\mathrm{3} \\ $$$${while}\:{if}\:\:{a}<\mathrm{0},\:{b}>\mathrm{3}\:. \\ $$
Commented by ajfour last updated on 19/May/17
Commented by ajfour last updated on 19/May/17
Commented by mrW1 last updated on 19/May/17
$$\mathrm{3}={ax}^{\mathrm{2}} +{b} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{3}−{b}}{{a}}>\mathrm{0} \\ $$$$\Rightarrow{a}>\mathrm{0}\:{and}\:{b}<\mathrm{3} \\ $$$${or} \\ $$$$\Rightarrow{a}<\mathrm{0}\:{and}\:{b}>\mathrm{3} \\ $$
Commented by ajfour last updated on 19/May/17
$${very}\:{straight},\:{thanks}\:. \\ $$