Question

I need help with solving this radical expression. I have been stuck on it for two days!
sqrt(8x-23)=4-x

Answer 1

The answer is 3, however I don't know how they got it.

Answer 2

what is 3

Answer 3

\[\sqrt{8x-23}=4-x\]
\[(\sqrt{8x-23})^2=(4-x)^2\]
That gets rid of the square root, and then you need to FOIL the right side. Can you do that for me?

Answer 4

(4-x)(4-x)=?

Answer 5

8x-23=16+x^2

Answer 6

or, 8x-23= 16-4x-4x=x^2

Answer 7

\[8x-23=x^2-8x+16\]
\[-x^2+16x-39=0\]
Then you should be able to factor and solve for the answer.
Or quadratic equation.

Answer 8

** or, 8x-23= 16-4x-4x=x^2 ***
I assume that last = is a typo ? It should be +

8x-23= 16-4x-4x+x^2
on the right side, -4x - 4x combines to make -8x
8x-23= 16-8x+x^2

switch sides (because that is how I am used to looking at these equations)
(and put in standard order)
x^2 - 8x +16 = 8x -23

add -8x +23 to both sides so that everything is on the left side
x^2 -8x -8x +16 +23 = 8x-23-8x +23

simplify
x^2 -16x +39 = 0

can we factor this ?
list pairs of factor so 39
1,39
3,13
look for a pair that add to 16. the 2nd pair work. make both - because we want -16

(x-3)(x-13)= 0

can you finish ?

remember to check the solution in the original equation, because the square root could cause troubles if you take the square root of a negative number

Answer 9

also, for this type equation
sqrt(8x-23)=4-x

you only take the + square root (the principal root)
that means you exclude any x values that make the right side negative ... because the left side is assumed to always be positive.

Answer 10

i did have a typo on the last one. I think I must have completely failed the quadratic equations section. that is where I lost it.

Answer 11

so since one of the blocks needs to be zero, (x-3)or (x-13) to equal zero I find that 3 is the only one that works in the original equation. sqrt of 8*3-23=4-3
1=1
the other solution of 13 does not work since 9 does not equal -9
THANKS!