Solving Conics: Find the x-Intercept

[emohunter7]

Conics! Help Please!

okay i think i have solved this correctly...still a little unsure though...
y=x^2-6x+3----find x intercept
first i used complete the square---- y-3=(x-3)^2
then i solved for x---- 3+(y-3)^(1/2)=x
then i plugged 0 in for y and got 3+(3^(1/2)) and 3-(3^(1/2))
is that correct??

 

[exk]

x intercept is where y=0.

 

[emohunter7]

yes i know that is why i made the equation equal to x and plugged 0 in for y

 

[exk]

why not just 0=x^2-6x+3?

 

[emohunter7]

you get the same answer i was just taught to solve for x first...so is my answer right or wrong??

 

[exk]

Try applying the quadratic formula to x^2-6x+3=0 and see if you get the same answer. I certainly don't.

Following your work:
then i solved for x---- 3+(y-3)^(1/2)=x
then i plugged 0 in for y and got 3+(3^(1/2)) and 3-(3^(1/2))

if you plug in 0 for y in 3+(y-3)^(1/2)=x you most certainly don't get 3+(3)^(1/2)=x, but 3+(-3)^(1/2)=x, which is not the answer.

The reason is that you didn't complete the square correctly: y=x^2-6x+3 --> y=x^2-6x+3+6-6 --> y=(x-3)^2-6 --> y+6=(x-3)^2

 

[emohunter7]

okay i see where i made my mistake..so the correct answer is x=3+(6^(1/2)) right?

 

[exk]

well yes and x=3-6^.5 also (remember there are 2 roots for a quadratic)

 

[emohunter7]

okay thank you for your help :)