Question

Find the number of real solutions of
\(\Large \frac{x^4-20x^3+150x^2-500x+625}{x-5}=0\)

Answer 1

ok hmmm lets see!!

Answer 2

see if x-5 is factorable first

Answer 3

because x=5 might infact be defined

Answer 4

yes

Answer 5

\[\large{x^4-20x^3+150x^2-500x+625}\]
\[\large{ = x^3(x-5) - 15x^2(x-5) + 75x(x-5)}\]
\[\large{ - 125(x-5)}\]

\[\large{= (x-5)(x^3-15x^2+75x-125)}\]

Answer 6

oops x=5 is good :)

Answer 7

infact its good again for the x^3 func too

Answer 8

Actually it is very very good. Its finally (x-5)^4

Answer 9

right so answer is (x-5)^3

Answer 10

5 is incorrect

Answer 11

\[(x-5)(x^3-15x^2+75x-125)\]

\[= (x-5)[x^2(x-5) - 10x(x-5) + 25(x-5)]\]

\[= (x-5)^2[x^2-10x+25]\]

\[= (x-5)^2(x-5)^2\]

\[=(x-5)^4\]

Answer 12

No real solution exists

Answer 13

Because we can only cancel out (x-5) in denominator and numerator if \(x\ne 5\)/

Answer 14

oh makes sense

Answer 15

Now our equation would become:

\[(x-5)^3 = 0 ;~~~~x\ne 5\]

\[\implies x = 5 ;~~~~~~~x\ne 5\]

Thus, no solution

Answer 16

then what happens at x=5?? do we say a hole ?

Answer 17

Yep

Answer 18

We get a 0/0 form

Answer 19

Function is non continuous at x = 5

Answer 20

quessioon tho

Answer 21

So, here are no real solutions and the answer is \(\Large 0\) :)

Answer 22

Yep :)

Answer 23

supposed u have
x^2=0

now i can multiply x/x * x^2=0 does this mean.., for x=0 its undefined now

Answer 24

Actually x/x would make your solution wrong

Answer 25

soo why cant this just be a (x-5)^3 cube shifted 5 to the right

Answer 26

how come

Answer 27

is it because u cant say some number over some number =1
since 0/0 and inf/inf or 0/inf is undefined

Answer 28

See, using x/x is possible only when \(x \ne 0\)

Answer 29

hmm ok

Answer 30

goood to knoww

Answer 31

For example taking square roots on LHS and RHS is possible only when you are sure that none of them is negative or we would gets stuck :)

Answer 32

(x^2-6) = -15

If I take square roots on both sides in this step, I am making a wrong step and I would get no solution

Answer 33

But if I solve it a little bit :

x^2 = 9

At this step taking square root is correct and permissible. I would get 2 solutions 3, -3

Answer 34

So, from one step we were getting no solutions but from the other we are getting 2 solutions. Only because we were using a wrong operation :)