Answers Integrated Math 2A - Mr. Katykhin 26Unit 2 Test

Unit 3 Test - Submit your Show Your Thinking into Canvas Answers

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Solve x2 + 6x = 7 by completing the square. Which is the solution set of the equation?

{

{–7, 1}

Option

{

{1, 7}

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Solve x2 – 8x = 20 by completing the square. Which is the solution set of the equation?

{

{–22, 14}

{

{–14, 22}

{

{–10, 2}

{

{–2, 10}

Solve x2 − 24x = −80 by completing the square.What is the solution set of the equation?

{

{2, 40}

{

{4, 20}

{

{5, 16}

{

{8, 10}

Using the quadratic formula to solve 11x2 – 4x = 1, what are the values of x?

StartFraction 2 Over 11 EndFraction plus-or-minus StartFraction StartRoot 15 EndRoot Over 11 EndFraction

StartFraction 2 Over 11 EndFraction plus-or-minus StartFraction 2 StartRoot 15 EndRoot Over 11 EndFraction

StartFraction 2 Over 11 EndFraction plus-or-minus StartFraction StartRoot 7 EndRoot Over 11 EndFraction

StartFraction 2 Over 11 EndFraction plus-or-minus StartFraction StartRoot 7 EndRoot i Over 11 EndFraction

Use the zero product property to find the solutions to the equation x2 – x – 6 = 0.

x = –3 or x = –2

x = –3 or x = 2

x = –2 or x = 3

x = 2 or x = 3

Which steps could be part of the process in algebraically solving the system of equations, y + 5x = x2 + 10 and y = 4x – 10? Select two options.

y = x2 + 5x + 10

y + 5x = x2 + 10 + 4x – 10

0 = x2 – 9x

0 = x2 – 9x + 20

One x-value of a solution to the system is 4.

If the discriminant of a quadratic equation is 4, which statement describes the roots?

There are two complex roots.

There are two real roots.

There is one real root.

There is one complex root.

Mischa wrote the quadratic equation 0=-x2+4x-7 in standard form. If a = -1, what is the value of c in her equation?

c=-7

c=-1

c=4

c=7

For which value of m does the graph of y = 18x2 + mx + 2 have exactly one x-intercept?

Which graph shows a system of equations with a solution at (2, –1)?

Option

What values of b satisfy 4(3b + 2)2 = 64?

b = and b = –2

b = 2 and b =

b = and b = 3

b = 2 and b =

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