Question

Solve for all real values of x :
(x^2 - 13x + 40)(x^2 - 13x + 42)/√(x^2 - 12x + 35)

Answer 1

is that a square root?

Answer 2

How can you solve that? It is not an equation.

Answer 3

yes

Answer 4

is it suppose to equal 0?

Answer 5

yes

Answer 6

@Mertsj its equal to 0 sorry

Answer 7

do you know how to factor?

Answer 8

i did that but that is as far as i got

Answer 9

so what do you have after factoring?

Answer 10

Consider the numerator. The only way a fraction can be 0 is if the numerator is 0 and the denominator is not 0. So set the numerator equal to 0. Then factor the numerator. Set each factor equal to 0. Solve each equation. Check to make sure that no answer causes the denominator to be 0.

Answer 11

@nikato (x-8)(x-5)(x-7)(x-6)/√(x-7)(x-5)

Answer 12

if you have
\[\frac{ (x-8)(x-5)(x-7)(x-6)}{\sqrt{(x-7)(x-5)} }=0\]
the top will be zero if any of the 4 factors is 0.
for example if x-8=0 (which of course means x=8), the top will be
0 * other stuff= 0

so x=8 will be one answer *as long as the bottom is also not zero*
as you can see, with x=8, the bottom is sqr(1*3) which is not zero. So x=8 is a solution

try the others.

Answer 13

so can i use 5 since it is x-5

Answer 14

what numbers in the denominator will cause it to become undefined?

Answer 15

yes, x=5 makes the top =0
but if the bottom is also 0 when x=5, you get 0/0 which is undefined, so not allowed

Answer 16

how are you determining that? did you just plug in 5 for x??

Answer 17

yes, just replace x with 5 and see what you get

Answer 18

in case it is not clear, if you have
(x-8)(x-5)(x-7)(x-6)=0
that can be broken down into 4 separate possibilities
x-8=0
x-5=0
x-7=0
x-6=0

those are the four ways to make the top equal to zero.
however, you have to reject any x value that ALSO causes the bottom to be zero
because we do not allow 0/0 as an answer.

Answer 19

so only 6 and 8 work right?

Answer 20

yes, exactly

Answer 21

are those all the zeros?