To determine if the quadratic equations have two real number solutions, we need to analyze their discriminants. The discriminant [tex]\( \Delta \)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
A quadratic equation will have:
- Two distinct real solutions if [tex]\( \Delta > 0 \)[/tex]
- One real solution (repeated) if [tex]\( \Delta = 0 \)[/tex]
- No real solutions if [tex]\( \Delta < 0 \)[/tex]
Let's analyze the discriminants of each given quadratic equation.
1. Equation: [tex]\( 0 = 2x^2 - 7x - 9 \)[/tex]
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -7 \)[/tex]
- [tex]\( c = -9 \)[/tex]
[tex]\[
\Delta = (-7)^2 - 4(2)(-9) = 49 + 72 = 121
\][/tex]
Since [tex]\( \Delta = 121 > 0 \)[/tex], this equation has two real solutions.
2. Equation: [tex]\( 0 = x^2 - 4x + 4 \)[/tex]
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -4 \)[/tex]
- [tex]\( c = 4 \)[/tex]
[tex]\[
\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0
\][/tex]
Since [tex]\( \Delta = 0 \)[/tex], this equation has one real solution (repeated).
3. Equation: [tex]\( 0 = 4x^2 - 3x - 1 \)[/tex]
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = -3 \)[/tex]
- [tex]\( c = -1 \)[/tex]
[tex]\[
\Delta = (-3)^2 - 4(4)(-1) = 9 + 16 = 25
\][/tex]
Since [tex]\( \Delta = 25 > 0 \)[/tex], this equation has two real solutions.
4. Equation: [tex]\( 0 = x^2 - 2x - 8 \)[/tex]
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]
- [tex]\( c = -8 \)[/tex]
[tex]\[
\Delta = (-2)^2 - 4(1)(-8) = 4 + 32 = 36
\][/tex]
Since [tex]\( \Delta = 36 > 0 \)[/tex], this equation has two real solutions.
5. Equation: [tex]\( 0 = 3x^2 + 5x + 3 \)[/tex]
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = 5 \)[/tex]
- [tex]\( c = 3 \)[/tex]
[tex]\[
\Delta = (5)^2 - 4(3)(3) = 25 - 36 = -11
\][/tex]
Since [tex]\( \Delta = -11 < 0 \)[/tex], this equation has no real solutions.
Therefore, the quadratic equations that have two real number solutions are:
[tex]\[
0 = 2x^2 - 7x - 9
\][/tex]
[tex]\[
0 = 4x^2 - 3x - 1
\][/tex]
[tex]\[
0 = x^2 - 2x - 8
\][/tex]
The equations [tex]\( 0 = x^2 - 4x + 4 \)[/tex] and [tex]\( 0 = 3x^2 + 5x + 3 \)[/tex] do not have two real number solutions.
