Select all the correct equations.
Which equations have no real solution but have two complex solutions?
3x^2- 5x= -8
2x^2= 6x – 5
12x= 9x^2 + 4
-x^2– 10x = 34
3x
2
−5x=−8
3x
2
−5x+8=0
This equation has the next form:
{ax}^{2} + bx + c = 0ax
2
+bx+c=0
To find if the equation has two complex solutions we have to check if the discriminant is negative, as follows:
\begin{gathered} {b}^{2} - 4ac \\ ( { - 5})^{2} - 4 \: . \: 3 \: . \: 8 = 25 - 96 = - 71 < 0\end{gathered}
b
2
−4ac
(−5)
2
−4.3.8=25−96=−71<0
Then, the first case has two complex solutions.
In the second case,
\begin{gathered} {2x}^{2} = 6x - 5 \\ {2x}^{2} - 6x + 5 = 0\end{gathered}
2x
2
=6x−5
2x
2
−6x+5=0
The discriminant in this case is:
( { - 6})^{2} - 4 \: . \: 2 \: . \: 5 = 36 - 40 = - 4 < 0(−6)
2
−4.2.5=36−40=−4<0
Then, the second case has two complex solutions.
In the third case,
\begin{gathered}12x = {9x}^{2} + 4 \\ { - 9x}^{2} + 12x - 4 = 0\end{gathered}
12x=9x
2
+4
−9x
2
+12x−4=0
The discriminant in this case is:
{12}^{2} - 4 \: . \: ( - 9) \: . \: ( - 4) = 144 - 144 = 012
2
−4.(−9).(−4)=144−144=0
Then, the third case has two real solutions.
In the fourth case,
\begin{gathered} { - x}^{2} - 10x = 34 \\ { - x}^{2} - 10x - 34 = 0\end{gathered}
−x
2
−10x=34
−x
2
−10x−34=0
The discriminant in this case is:
( { - 10})^{2} - 4 \: . \: ( - 1) \: . \: ( - 34) = 100 - 136 = - 36 < 0(−10)
2
−4.(−1).(−34)=100−136=−36<0
Then, the fourth case has two complex solutions.
