Question

PLZ HELL FAST WILL GIVE BRAINLIEST

Answer 1

${3x}^{2} - 5x = - 8 \\ {3x}^{2} - 5x + 8 = 0$

This equation has the next form:

${ax}^{2} + bx + c = 0$

To find if the equation has two complex solutions we have to check if the discriminant is negative, as follows:

${b}^{2} - 4ac \\ ( { - 5})^{2} - 4 \: . \: 3 \: . \: 8 = 25 - 96 = - 71 < 0$

Then, the first case has two complex solutions.

In the second case,

${2x}^{2} = 6x - 5 \\ {2x}^{2} - 6x + 5 = 0$

The discriminant in this case is:

$( { - 6})^{2} - 4 \: . \: 2 \: . \: 5 = 36 - 40 = - 4 < 0$

Then, the second case has two complex solutions.

In the third case,

$12x = {9x}^{2} + 4 \\ { - 9x}^{2} + 12x - 4 = 0$

The discriminant in this case is:

${12}^{2} - 4 \: . \: ( - 9) \: . \: ( - 4) = 144 - 144 = 0$

Then, the third case has two real solutions.

In the fourth case,

${ - x}^{2} - 10x = 34 \\ { - x}^{2} - 10x - 34 = 0$

The discriminant in this case is:

$( { - 10})^{2} - 4 \: . \: ( - 1) \: . \: ( - 34) = 100 - 136 = - 36 < 0$

Then, the fourth case has two complex solutions.

Answer 2

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.