AnswersCR - Algebra 2 26-27 - S1Quadratic in Form Polynomials

Graphing Polynomial Functions Answers

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Which quadratic equation is equivalent to (x2 – 1)2 – 11(x2 – 1) + 24 = 0?

A
u2 – 11u + 24 = 0 where u = (x2 – 1)
B
(u2)2 – 11(u2) + 24 where u = (x2 – 1)
C
u2 + 1 – 11u + 24 = 0 where u = (x2 – 1)
D
(u2 – 1)2 – 11(u2 – 1) + 24 where u = (x2 – 1)
3

What is the solution of the equation (x – 5)2 + 3(x – 5) + 9 = 0? Use u substitution and the quadratic formula to solve.

A
x = StartFraction negative 3 plus-or-minus 3 i StartRoot 3 EndRoot Over 2 EndFraction
Option A
B
x = StartFraction 7 plus-or-minus 3 i StartRoot 3 EndRoot Over 2 EndFraction
Option B
C
x = 2
D
x = 8
5

Which equation is quadratic in form?

A
4(x – 2)2 + 3x – 2 + 1 = 0
B
8x5 + 4x3 + 1 = 0
C
10x8 + 7x4 + 1 = 0
D
9x16 + 6x4 + 1 = 0

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