Question

I have five Algebra II questions... Can someone help me? I'm posting them shortly...

Answer 1

@phi @amistre64 @AccessDenied

Answer 2

Solve the equations for x. Identify any extraneous solutions. Show all work.

Answer 3

If somebody can show me one or two of them, I might be able to get the others by myself...

Answer 4

to solve the square root, you need to square both sides and then you have

8x+1 = 25

Answer 5

from there can you solve for x

Answer 6

what's an extraneous solution?

Answer 7

I got x=3

Answer 8

correct

Answer 9

there aren't any other answers, right?

Answer 10

the next problem, first you subtract 5 from both sides, then you square

Answer 11

x=43?

Answer 12

I get the following, you have

sqrt(x-7) = 11- 5
sqrt(x-7) = 6

Answer 13

now square, then you get

x - 7 = 36

Answer 14

now solve for x

Answer 15

x=43

Answer 16

correct, you were right all alone

Answer 17

are there any extraneous solutions?

Answer 18

if you get more than 1 possible result, and some test false; the false ones are called extras (extranuous)

Answer 19

oh okay

Answer 20

For #3, do you start by dividing both sides by -4?

Answer 21

number 5 has the possibilty of getting some extras since you would have to square an x and solve a quadratic

Answer 22

yes, divide by -4 first

Answer 23

i might want to amend my thought of extras :)

Answer 24

x=16?

Answer 25

all false solutions are extras; which means that there may be no real solutions

Answer 26

x=16 is the only thing that might have a chance at working; but the test is false

Answer 27

so it's the only solution but it's extraneous?

Answer 28

correct

-4(5) = 20, there is no real solution

Answer 29

but there is one extraneous?

Answer 30

correct

Answer 31

Number 4 is x=8?

Answer 32

correct

Answer 33

I was just double checking my work and it looks like Number 3 has a solution... What did I do wrong?

Answer 34

Plugging in 16 for X

-4sqrt(16+9)=20
sqrt(16+9)=-5
16+9=25
25=25

Answer 35

@amistre64

Answer 36

I don't understand number 5 at all

Answer 37

first subtract the 4 and you get the following

sqrt(x+2) = x - 4

Answer 38

then when you square, you have

x+ 2 = (x-4)^2

Answer 39

then you solve as follows;

x+2 = (x-4)(x-4)
x+2 = x^2-4x-4x+16
x+2 = x^2-8x+16

Answer 40

can you take it from there

Answer 41

@ivettef365 what do you do with the x^2?

Answer 42

ok, this is what you do

x+2-x-2 = x^2-8x+16-x-2
0 = x^2 -9x+14

Answer 43

now x^2-9x+14

(x-7)(x-2) = 0

Answer 44

so now to find value of x,

x-7 = 0
x = 7

and

x-2 = 0
x = 2