To determine which equations have no real solutions but have two complex solutions, we need to analyze each equation separately to see if it yields complex solutions. An equation has complex solutions when the discriminant (the part under the square root in the quadratic formula [tex]\(\Delta = b^2 - 4ac\)[/tex]) is negative. If [tex]\(\Delta\)[/tex] is negative, the quadratic equation does not cross the x-axis and thus has no real solutions, only complex ones.
Let's examine each equation:
1. [tex]\(3x^2 - 5x = -8\)[/tex]
This can be rewritten as:
[tex]\[
3x^2 - 5x + 8 = 0
\][/tex]
Here, [tex]\(a = 3\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = 8\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac = (-5)^2 - 4(3)(8) = 25 - 96 = -71
\][/tex]
Since [tex]\(\Delta < 0\)[/tex], this equation has no real solutions but has two complex solutions.
2. [tex]\(2x^2 = 6x - 5\)[/tex]
This can be rewritten as:
[tex]\[
2x^2 - 6x + 5 = 0
\][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = 5\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4
\][/tex]
Since [tex]\(\Delta < 0\)[/tex], this equation also has no real solutions but has two complex solutions.
3. [tex]\(12x = 9x^2 + 4\)[/tex]
This can be rewritten as:
[tex]\[
9x^2 - 12x + 4 = 0
\][/tex]
Here, [tex]\(a = 9\)[/tex], [tex]\(b = -12\)[/tex], and [tex]\(c = 4\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac = (-12)^2 - 4(9)(4) = 144 - 144 = 0
\][/tex]
Since [tex]\(\Delta = 0\)[/tex], this equation has exactly one real solution (a repeated real root), not complex solutions.
4. [tex]\(-x^2 - 10x = 34\)[/tex]
This can be rewritten as:
[tex]\[
-x^2 - 10x - 34 = 0
\][/tex]
Or multiplying by [tex]\(-1\)[/tex] to make it standard form:
[tex]\[
x^2 + 10x + 34 = 0
\][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 34\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac = 10^2 - 4(1)(34) = 100 - 136 = -36
\][/tex]
Since [tex]\(\Delta < 0\)[/tex], this equation has no real solutions but has two complex solutions.
Thus, 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]
