AnswersAZ-College Prep Math BThe Quadratic Formula

The Quadratic Formula Answers

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If x = –3 is the only x-intercept of the graph of a quadratic equation, which statement best describes the discriminant of the equation?

A
The discriminant is negative.
B
The discriminant is –3.
C
The discriminant is 0.
D
The discriminant is positive.
3

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x?

A
StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoot Over 12 EndFraction
Option A
B
StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction
Option B
C
StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction
Option C
D
StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction
Option D
4

Using the quadratic formula to solve x2 = 5 – x, what are the values of x?

A
StartFraction negative 1 plus-or-minus StartRoot 21 EndRoot Over 2 EndFraction
Option A
B
StartFraction negative 1 plus-or-minus StartRoot 19 EndRoot i Over 2 EndFraction
Option B
C
StartFraction 5 plus-or-minus StartRoot 21 EndRoot Over 2 EndFraction
Option C
D
StartFraction 1 plus-or-minus StartRoot 19 EndRoot i Over 2 EndFraction
Option D
8

Using the quadratic formula to solve 7x2 – x = 7, what are the values of x?

A
StartFraction 1 plus-or-minus StartRoot 195 EndRoot i Over 14 EndFraction
Option A
B
StartFraction 1 plus-or-minus StartRoot 197 EndRoot Over 14 EndFraction
Option B
C
StartFraction 1 plus-or-minus StartRoot 195 EndRoot Over 14 EndFraction
Option C
D
StartFraction 1 plus-or-minus StartRoot 197 EndRoot i Over 14 EndFraction
Option D
9

Which equation has the solutions ?

Question illustration
A
3x2 – 5x + 7 = 0
B
3x2 – 5x – 1 = 0
C
3x2 – 10x + 6 = 0
D
3x2 – 10x – 1 = 0
10

Which equation shows the quadratic formula used correctly to solve 5x2 + 3x – 4 = 0 for x?

A
x = StartFraction negative 3 plus-or-minus StartRoot (3) squared minus 4 (5) (negative 4) EndRoot Over 2 (5) EndFraction
Option A
B
x = StartFraction 3 plus-or-minus StartRoot (3) squared + 4 (5) (negative 4) EndRoot Over 2 (5) EndFraction
Option B
C
x = StartFraction 3 plus-or-minus StartRoot (3) squared minus 4 (5) (negative 4) EndRoot Over 2 (5) EndFraction
Option C
D
x = StartFraction negative 3 plus-or-minus StartRoot (3) squared + 4 (5) (negative 4) EndRoot Over 2 (5) EndFraction
Option D

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