AnswersNY-Algebra I Term 2 HillcrestSolving Linear-Quadratic Systems

Solving Linear-Quadratic Systems — Unit test Answers

25 verified answers
5

A function g(x) has x-intercepts at (, 0) and (6, 0). Which could be g(x)?

Question illustration
A
g(x) = 2(x + 1)(x + 6)
B
g(x) = (x – 6)(2x – 1)
C
g(x) = 2(x – 2)(x – 6)
D
g(x) = (x + 6)(x + 2)
6

Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation?

A
(negative 6 minus StartRoot 51 EndRoot comma negative 6 + StartRoot 51 EndRoot)
Option A
B
(negative 6 minus StartRoot 21 EndRoot comma negative 6 + StartRoot 21 EndRoot)
Option B
C
(6 minus StartRoot 51 EndRoot comma 6 + StartRoot 51 EndRoot)
Option C
D
(6 minus StartRoot 21 EndRoot comma 6 + StartRoot 21 EndRoot)
Option D
7

What are the values of a, b, and c in the quadratic equation 0 = 5x – 4x2 – 2?

A
a = 5, b = 4, c = 2
B
a = 5, b = –4, c = –2
C
a = –4, b = 5, c = –2
D
a = 4, b = –5, c = –2
10

Which equations can be used to find the lengths of the legs of the triangle? Select three options.

A
0.5(x)(x + 2) = 24
B
x(x + 2) = 24
C
x2 + 2x – 24 = 0
D
x2 + 2x – 48 = 0
E
x2 + (x + 2)2 = 100
11

What is the first step in solving the quadratic equation -5x2+8=133?

A
taking the square root of both sides of the equation
B
subtracting 8 from both sides of the equation
C
squaring both sides of the equation
D
adding 8 to both sides of the equation
13

Which shows the correct substitution of the values a, b, and c from the equation –2 = –x + x2 – 4 into the quadratic formula?Quadratic formula: x =

Question illustration
A
x = StartFraction negative (negative 1) plus or minus StartRoot (negative 1) squared minus 4 (1)(negative 4) EndRoot Over 2(1) EndFraction
Option A
B
x = StartFraction negative 1 plus or minus StartRoot 1 squared minus 4 (negative 1)(negative 4) EndRoot Over 2(negative 1) EndFraction
Option B
C
x = StartFraction negative (negative 1) plus or minus StartRoot (negative 1) squared minus 4 (1)(negative 4) EndRoot Over 2(1) EndFraction
Option C
D
x = StartFraction negative (negative 1) plus or minus StartRoot (negative 1) squared minus 4 (1)(negative 2) EndRoot Over 2(1) EndFraction
Option D
17

Zacharias is using the quadratic formula to solve the equation 0 = –2x2 + 5x – 3. He begins by substituting as shown.Quadratic formula: x = Substitution:

Question illustration
A
The –5 should be 5.
B
The 52 should be –52.
C
The 2 in the numerator should be –2.
D
The 2 in the denominator should be –2.
20

Which shows the correct substitution of the values a, b, and c from the equation 0 = 4x2 + 2x – 1 into the quadratic formula below?Quadratic formula: x =

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

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