Eloise started to solve a radical equation in this way: square root of negative 2x plus 1 − 3 = x
square root of negative 2x plus 1 − 3 + 3 = x + 3
square root of negative 2x plus 1 = x + 3
square root of negative 2x plus 1 − 1 = x + 3 − 1
square root of negative 2 x = x + 2
(square root of negative 2 x)2 = (x − 4)2
−2x = x2 − 8x + 16
−2x + 2x = x2 + 8x + 16 + 2x
0 = x2 + 10x + 16
0 = (x + 2)(x + 8)

x + 2 = 0 x + 8 = 0
x + 2 − 2 = 0 − 2 x + 8 − 8 = 0 − 8
x = −2 x = −8

Both solutions are extraneous because they don't satisfy the original equation.

What error did Eloise make?

A) She added 2x after squaring both sides.
B) She subtracted 1 before squaring both sides.
C) She factored x2 + 10x + 16 incorrectly.
D) She did not check for extraneous solutions.

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

3030 ur welcome

step-by-step explanation:

i have no idea just need points sorry

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.