Conics Help Please

  • Thread starter emohunter7
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19
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Main Question or Discussion Point

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??
 

Answers and Replies

exk
119
0
x intercept is where y=0.
 
19
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yes i know that is why i made the equation equal to x and plugged 0 in for y
 
exk
119
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why not just 0=x^2-6x+3?
 
19
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you get the same answer i was just taught to solve for x first.....so is my answer right or wrong??
 
exk
119
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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
 
19
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okay i see where i made my mistake..so the correct answer is x=3+(6^(1/2)) right???
 
exk
119
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well yes and x=3-6^.5 also (remember there are 2 roots for a quadratic)
 
19
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okay thank you for your help :)
 

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