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Find the real roots of an equation

I managed to get y = -6 and 1 but don't know how to get x :/

Normally you just square root/square with these types of questions so I'm not use to this one.

http://imgur.com/OenWCeo
Original post by PeaceTreaty
I managed to get y = -6 and 1 but don't know how to get x :/

Normally you just square root/square with these types of questions so I'm not use to this one.

http://imgur.com/OenWCeo

Looks like there's be 4 values for x (unless I'm mistaken). I'm not surprised as it is a quartic.
y=(x+2)2y=(x+2)^2 so 6=(x+2)2-6=(x+2)^2 and 1=(x+2)21=(x+2)^2. Actually, just 2 real roots looking at that. Although you'll also have 2 complex ones.
Original post by PeaceTreaty
I managed to get y = -6 and 1 but don't know how to get x :/

Normally you just square root/square with these types of questions so I'm not use to this one.

http://imgur.com/OenWCeo


You know that y=(x+2)20y=(x+2)^2 \geq 0

You have two y values, one positive and one negative. Which one can you scrap?
Reply 3
Original post by keromedic
Looks like there's be 4 values for x (unless I'm mistaken). I'm not surprised as it is a quartic.
y=(x+2)2y=(x+2)^2 so 6=(x+2)2-6=(x+2)^2 and 1=(x+2)21=(x+2)^2. Actually, just 2 real roots looking at that. Although you'll also have 2 complex ones.



Original post by Slowbro93
You know that y=(x+2)20y=(x+2)^2 \geq 0

You have two y values, one positive and one negative. Which one can you scrap?


I figured it out (x+2=+\-1) thanks.
Original post by PeaceTreaty
I figured it out (x+2=+\-1) thanks.


No problem :h:

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