Answers

2016-04-21T21:19:55+05:30
ux^2+vx+w=0\\\Delta=v^2-4uw\\x_1=\frac{-v-\sqrt\Delta}{2u}\\x_2=\frac{-v+\sqrt\Delta}{2u}\\\\x^2+ 5x-(a^2+a-6)=0\\\Delta=5^2-(-4(<span>a^2+a-6))=25+4a^2+4a-24=4a^2+4a+1\\\Delta_{\Delta}=4^2-4\times4=0\\a_{1,2}=-0.5
Δ must be ≥ 0, because expression in √ must be no negative, so a ≥ -0.5:
x_1=\frac{-5-\sqrt{4a^2+4a+1}}{2}\\<span>x_2=\frac{-5+\sqrt{4a^2+4a+1}}{2}

If a > -0.5  equation has 2 different real solutions,
if a = -0.5  equation has 1 dual real solution,
if a < =0.5 equation has no real solution (it has solutions, which are complex numbers).

0