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a-b-c-R-a-0-and-the-quadratic-equation-ax-2-bx-c-0-has-no-real-roots-then-

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Question Number 34425 by Vijay kumar prasad last updated on 06/May/18
a, b, c ∈ R, a≠0 and the quadratic equation ax^2 +bx+c=0 has no real roots, then
$${a},\:{b},\:{c}\:\in\:{R},\:{a}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real} \\ $$$$\mathrm{roots},\:\mathrm{then} \\ $$
Answered by MJS last updated on 06/May/18
x=((−b±(√(b^2 −4ac)))/(2a)) b^2 −4ac<0 ⇒ no real roots b^2 −4ac=0 ⇒ exactly one real root b^2 −4ac>0 ⇒ two real roots
$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{roots} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}\:\Rightarrow\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{0}\:\Rightarrow\:\mathrm{two}\:\mathrm{real}\:\mathrm{roots} \\ $$
Answered by Rio Mike last updated on 06/May/18
b^2 − 4ac = 0
$$\:{b}^{\mathrm{2}} \:−\:\mathrm{4}{ac}\:=\:\mathrm{0} \\ $$
Commented by MJS last updated on 06/May/18
in this case you have 1 real root
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{you}\:\mathrm{have}\:\mathrm{1}\:\mathrm{real}\:\mathrm{root} \\ $$

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