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 .


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

$F o r t w o r e a l a n d d i s t i n c t$ $s o l u t i o n s t o :$ $y = 3$ $y = {a x}^{2} + b$ $A s c a n b e s e e n f r o m g r a p h i n$ $c o m m e n t b e l o w,$ $i f a > 0, b < 3$ $w h i l e i f a < 0, b > 3 .$
Commented byajfour last updated on 19/May/17

Commented byajfour last updated on 19/May/17

Commented bymrW1 last updated on 19/May/17

$3 = {a x}^{2} + b$ $x^{2} = \frac{3 - b}{a} > 0$ $\Rightarrow a > 0 a n d b < 3$ $o r$ $\Rightarrow a < 0 a n d b > 3$
Commented byajfour last updated on 19/May/17

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