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Substitute the values for [tex]$a, b,$[/tex] and [tex]$c$[/tex] into [tex]$b^2 - 4ac$[/tex] to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two [tex]$x$[/tex]-intercepts.) Check all that apply.

A. [tex]$0 = 2x^2 - 7x - 9$[/tex]
B. [tex]$0 = x^2 - 4x + 4$[/tex]
C. [tex]$0 = 4x^2 - 3x - 1$[/tex]
D. [tex]$0 = x^2 - 2x - 8$[/tex]
E. [tex]$0 = 3x^2 + 5x + 3$[/tex]

Sagot :

GinnyAnsweridn
To determine which quadratic equations have two real solutions, we need to calculate the discriminant [tex]\( \Delta = b^2 - 4ac \)[/tex]. If the discriminant is greater than 0, the quadratic equation will have two distinct real solutions. Let's evaluate the discriminant for each equation:

1. [tex]\( 0 = 2x^2 - 7x - 9 \)[/tex]
- Coefficients: [tex]\( a = 2 \)[/tex], [tex]\( b = -7 \)[/tex], [tex]\( c = -9 \)[/tex]
- Discriminant: [tex]\( \Delta = (-7)^2 - 4(2)(-9) \)[/tex]
- Calculation: [tex]\( \Delta = 49 + 72 = 121 \)[/tex]
- Since [tex]\( \Delta > 0 \)[/tex], this equation has two real solutions.

2. [tex]\( 0 = x^2 - 4x + 4 \)[/tex]
- Coefficients: [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], [tex]\( c = 4 \)[/tex]
- Discriminant: [tex]\( \Delta = (-4)^2 - 4(1)(4) \)[/tex]
- Calculation: [tex]\( \Delta = 16 - 16 = 0 \)[/tex]
- Since [tex]\( \Delta = 0 \)[/tex], this equation does not have two real solutions (it has one real solution).

3. [tex]\( 0 = 4x^2 - 3x - 1 \)[/tex]
- Coefficients: [tex]\( a = 4 \)[/tex], [tex]\( b = -3 \)[/tex], [tex]\( c = -1 \)[/tex]
- Discriminant: [tex]\( \Delta = (-3)^2 - 4(4)(-1) \)[/tex]
- Calculation: [tex]\( \Delta = 9 + 16 = 25 \)[/tex]
- Since [tex]\( \Delta > 0 \)[/tex], this equation has two real solutions.

4. [tex]\( 0 = x^2 - 2x - 8 \)[/tex]
- Coefficients: [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], [tex]\( c = -8 \)[/tex]
- Discriminant: [tex]\( \Delta = (-2)^2 - 4(1)(-8) \)[/tex]
- Calculation: [tex]\( \Delta = 4 + 32 = 36 \)[/tex]
- Since [tex]\( \Delta > 0 \)[/tex], this equation has two real solutions.

5. [tex]\( 0 = 3x^2 + 5x + 3 \)[/tex]
- Coefficients: [tex]\( a = 3 \)[/tex], [tex]\( b = 5 \)[/tex], [tex]\( c = 3 \)[/tex]
- Discriminant: [tex]\( \Delta = (5)^2 - 4(3)(3) \)[/tex]
- Calculation: [tex]\( \Delta = 25 - 36 = -11 \)[/tex]
- Since [tex]\( \Delta < 0 \)[/tex], this equation does not have two real solutions.

Based on these calculations, the quadratic equations that have two real number solutions are:

- Equation 1: [tex]\( 0 = 2x^2 - 7x - 9 \)[/tex]
- Equation 3: [tex]\( 0 = 4x^2 - 3x - 1 \)[/tex]
- Equation 4: [tex]\( 0 = x^2 - 2x - 8 \)[/tex]

Therefore, the quadratic equations with two real number solutions are equations 1, 3, and 4.
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