Question

i really need help

Answer 1

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Answer 2

following the directions in a)

x + a = sqrt(bx + c)

they tell you to plug in x = 7
they also tell you to plug in a number for a (we can start with a = 1 for now, and change it if needed)

this gives us

7 + 1 = sqrt(b*7 + c)

squaring this gives us

64 = 7b + c

so you just have to pick integers for b and c that makes this true

Answer 3

that make them = 64 true?

Answer 4

64 = 7b + c

64 = 7(9) + 1

Answer 5

good, so we can try b = 9 and c = 1 to see if we get a good equation with extraneous solutions

so x + a = sqrt(bx + c)
becomes

x + 1 = sqrt(9x + 1)

try squaring both sides of this equation, and solving for x

Answer 6

x^2+2x+1=9x + 1
subtracted 9x+1 to both

then factored which gives me a solution of
0 or 7

Answer 7

awesome

plugging x = 0 back into the original equation

x + 1 = sqrt(9x + 1)
0 + 1 = sqrt(9*0+1)
1 = 1

so x = 0 is not an extraneous solution (to be an extraneous solution, it must not work when you plug it back into the original equation)

but that's ok, we have fulfilled the requirement for parts a) and b) we can move on to part c) now

Answer 8

c) asks us to re-write the equation, picking different a, b, and c values such that there is an extraneous solution

I already did the calculations and found that a = 1, b = 7 and c = 15 will produce an extraneous solution

x + a = sqrt(bx + c)
becomes

x + 1 = sqrt(7*x+15)

plugging in x = 7 gives us

7 + 1 = sqrt(7*7 + 15)
8 = 8

Answer 9

now, going back and solving

x + 1 = sqrt(7*x+15)

gives us

(x+1)^2 = 7x + 15

x^2 + 2x + 1 = 7x + 15
x^2 - 5x - 14 = 0

gives us x = -2 and x = 7

Answer 10

however, x = -2 is extraneous because we can plug in x = -2 back into the equation and get

x + a = sqrt(bx + c)

-2 + 1 = sqrt(-2*7 + 15)
-1 =/= sqrt(1)
therefore a possible solution for part c) is

x + 1 = sqrt(7x + 15)

Answer 11

this is a very involved problem so feel free to ask me if anything was confusing or unclear

Answer 12

wow thank you i learned alot !