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Select all the correct equations.

Which equations have no real solution but have two complex solutions?

\begin{tabular}{|c|c|}
\hline
[tex]$3x^2 - 5x = -8$[/tex] & [tex]$2x^2 = 6x - 5$[/tex] \\
\hline
[tex]$12x = 9x^2 + 4$[/tex] & [tex]$-x^2 - 10x = 34$[/tex] \\
\hline
\end{tabular}

Sagot :

To determine which equations have no real solutions but have two complex solutions, we need to analyze the discriminant of each quadratic equation. For a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], the discriminant is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].

If the discriminant is less than 0 ([tex]\( \Delta < 0 \)[/tex]), the quadratic equation has two complex solutions.

Let's analyze each given equation:

1. [tex]\( 3x^2 - 5x = -8 \)[/tex]
[tex]\[ \Rightarrow 3x^2 - 5x + 8 = 0 \][/tex]
The discriminant is:
[tex]\[ \Delta = (-5)^2 - 4 \cdot 3 \cdot 8 = 25 - 96 = -71 \][/tex]
Since the discriminant is less than 0, this equation has two complex solutions.

2. [tex]\( 2x^2 = 6x - 5 \)[/tex]
[tex]\[ \Rightarrow 2x^2 - 6x + 5 = 0 \][/tex]
The discriminant is:
[tex]\[ \Delta = (-6)^2 - 4 \cdot 2 \cdot 5 = 36 - 40 = -4 \][/tex]
Since the discriminant is less than 0, this equation also has two complex solutions.

3. [tex]\( 12x = 9x^2 + 4 \)[/tex]
[tex]\[ \Rightarrow 9x^2 - 12x + 4 = 0 \][/tex]
The discriminant is:
[tex]\[ \Delta = (-12)^2 - 4 \cdot 9 \cdot 4 = 144 - 144 = 0 \][/tex]
Since the discriminant is equal to 0, this equation has a double root (one real solution) and not complex solutions.

4. [tex]\( -x^2 - 10x = 34 \)[/tex]
[tex]\[ \Rightarrow -x^2 - 10x - 34 = 0 \][/tex]
The discriminant is:
[tex]\[ \Delta = (-10)^2 - 4 \cdot (-1) \cdot (-34) = 100 - 136 = -36 \][/tex]
Since the discriminant is less than 0, this equation has two complex solutions.

Therefore, the equations that have no real solutions but have two complex solutions are:
[tex]\[ \begin{array}{|c|c|} \hline 3x^2 - 5x = -8 & 2x^2 = 6x - 5 \\ \hline -x^2 - 10x = 34 & \hline \end{array} \][/tex]