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Question Number 13359 by ajfour last updated on 19/May/17

For two real and distinct  solutions to :  y=3  y=ax^2 +b  As can be seen from graph in  comment below,  if a>0, b<3  while if  a<0, b>3 .

$${For}\:{two}\:{real}\:{and}\:{distinct} \\ $$ $${solutions}\:{to}\:: \\ $$ $${y}=\mathrm{3} \\ $$ $${y}={ax}^{\mathrm{2}} +{b} \\ $$ $${As}\:{can}\:{be}\:{seen}\:{from}\:{graph}\:{in} \\ $$ $${comment}\:{below}, \\ $$ $${if}\:{a}>\mathrm{0},\:{b}<\mathrm{3} \\ $$ $${while}\:{if}\:\:{a}<\mathrm{0},\:{b}>\mathrm{3}\:. \\ $$

Commented byajfour last updated on 19/May/17

Commented byajfour last updated on 19/May/17

Commented bymrW1 last updated on 19/May/17

3=ax^2 +b  x^2 =((3−b)/a)>0  ⇒a>0 and b<3  or  ⇒a<0 and b>3

$$\mathrm{3}={ax}^{\mathrm{2}} +{b} \\ $$ $${x}^{\mathrm{2}} =\frac{\mathrm{3}−{b}}{{a}}>\mathrm{0} \\ $$ $$\Rightarrow{a}>\mathrm{0}\:{and}\:{b}<\mathrm{3} \\ $$ $${or} \\ $$ $$\Rightarrow{a}<\mathrm{0}\:{and}\:{b}>\mathrm{3} \\ $$

Commented byajfour last updated on 19/May/17

very straight, thanks .

$${very}\:{straight},\:{thanks}\:. \\ $$

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