AnswersFL-1200710-Mathematics for College Algebra ASolving Quadratic Equations: Completing the Square (Continued)

Solving Quadratic Equations: Quadratic Formula Answers

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Mattie uses the discriminant to determine the number of zeros the quadratic equation 0 = 3x2 – 7x + 4 has. Which best describes the discriminant and the number of zeros?

A
The equation has one zero because the discriminant is 1.
B
The equation has one zero because the discriminant is a perfect square.
C
The equation has two zeros because the discriminant is greater than 0.
D
The equation has no zeros because the discriminant is not a perfect square.
3

Anderson uses the discriminant to correctly find the number of real solutions of the quadratic equation x2 + 4x + 8 = 0. Which explanation could Anderson provide?

Question illustration
A
The equation has no real number solutions because the discriminant is 0.
B
The equation has one real number solution because the discriminant is 0.
C
The equation has no real number solutions because the discriminant is less than 0.
D
The equation has two real number solutions because the discriminant is greater than 0.
5

What is the value of the discriminant of the quadratic equation -2x2=-8x+8, and what does its value mean about the number of real number solutions the equation has?

A
The discriminant is equal to 0, which means the equation has no real number solutions.
B
The discriminant is equal to 0, which means the equation has one real number solution.
C
The discriminant is equal to 128, which means the equation has no real number solutions.
D
The discriminant is equal to 128, which means the equation has two real number solutions.
6

A quadratic equation has a discriminant of 12. Which could be the equation?

A
0 = –x2 + 8x + 2
B
0 = 2x2 + 6x + 3
C
0 = –x2 + 4x + 1
D
0 = 4x2 + 2x + 1
9

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the values a, b, and c into the quadratic formula?Quadratic formula: x =

Question illustration
A
StartFraction 2 plus or minus StartRoot (negative 2) squared minus 4(1)(negative 3) EndRoot Over 2(1) EndFraction
Option A
B
StartFraction negative 2 plus or minus StartRoot (negative 2) squared minus 4(1)(negative 3) EndRoot Over 2(1) EndFraction
Option B
C
StartFraction 2 plus or minus StartRoot negative 2 squared minus 4(1)(negative 3) EndRoot Over 2(1) EndFraction
Option C
D
StartFraction negative 2 plus or minus StartRoot negative 2 squared minus 4(1)(negative 3) EndRoot Over 2(1) EndFraction
Option D

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