Question

how would i find the intercepts for the equation y=(x-1)√x+1 ?
the 1 is also under the square root

Answer 1

set x = 0 and solve for y to get the y intercept.

set y = 0 and solve for x to get the x intercept(s) or roots as they are often called

Answer 2

when i set y=0 how would i solve that?

Answer 3

\[0=\left( x-1 \right)\sqrt{x+1}=x \sqrt{x+1}-\sqrt{x+1}\rightarrow \sqrt{x+1}=x \sqrt{x+1}\]

do you know how to solve this? if so, thank your algebra teacher.
if not, your algebra teacher didn't do their job.

Answer 4

is it x=0?

Answer 5

plug it in... what do you get on the left? what do you get on the right? are they equal?

Answer 6

alright so i'm wrong and my algebra teacher didnt do their job, could you fill me in on what he didnt?

Answer 7

wait it's x=1, right?

Answer 8

wait nvm just ignore me haha

Answer 9

sure.
to solve this you need to square both sides. from there, you'll need to move everything over to one side and find the roots of the resulting polynomial. then you must check each root in the original equation because you can get what are called extraneous roots.
they are roots that you get from the solving process but don't actually work in the original equation.
take it step by step and i'll help you along.

Answer 10

there may be another solution too... hint, hint. wink, wink.

Answer 11

so whenever i square the right side will i get x^3 +1 or x^3 +x ? or neither, im prepared to be wrong again

Answer 12

sorry, i goofed a little.

we have two terms which are being multiplied. so either can be 0 and the whole thing is o

what makes (x-1) = 0?

what makes sqrt(x+1) = 0

those are your roots.

sorry for the sidetrack!

Answer 13

make sense?

Answer 14

i just realized i made the question wrong, the x should be squared under the square root in the original equation. hahaha

Answer 15

it's all the same except you'll never get that term to equal 0 unless you can use complex numbers.

Answer 16

Solve it step by step so at least i have a reference

Answer 17

so y = (x-1)sqrt(x^2+1) which means in order for y to be 0, either
x-1 = 0 or sqrt(x^2 + 1) = 0
the first one implies x = 1
the second one implies that x^2+1 = 0 => x^2 = -1 which no real number can satisfy this.

so the only root is x=1