Question

anyone know how to solve x^2+x=12 by completing the square?

Answer 1

I love completing the square! Would you like help with this?

Answer 2

Yes, i really would. I practiced it a while back but now i forgot

Answer 3

Ok, the first rule is that the coefficient in front of the x^2 has to be a 1. It is. Cool. Next rule is that you have the x terms on one side of the equal sign and the constants on the other. You do. Even cooler! Now we can complete the square. To do that, take half the coefficient on the x term and square it. The coefficient on the x term is 1. Half of 1 is 1/2 right? Squaring 1/2 what do you get?

Answer 4

1/4

Answer 5

Good. So now that you found 1/4, you add it in after the x term. BUT (and this is a big but) [lol] if you add it to one side of the equation, you have to add it to the other side. Because if you don't you won't have an equation anymore. You have to keep it in balance. So here's what you did so far:

Answer 6

that's the part i wasn't sure of: If i should add or subtract it

Answer 7

\[x ^{2}+1x+\frac{ 1 }{ 4 }=12+\frac{ 1 }{ 4 }\].

Answer 8

What you create when you halve the x term and square it is a perfect binomial, which is the whole point of this process. The perfect binomial you created is \[(x+\frac{ 1 }{ 2 })^{2}\]Because you halved the x term, got 1/2 and then squared it right?

Answer 9

Now you need to add the 12 and the 1/4. LCD is 4, so\[(x+\frac{ 1 }{ 2 })^{2}=\frac{ 48 }{ 4 }+\frac{ 1 }{ 4 }\]That gives you\[(x+\frac{ 1 }{ 2 })^{2}=\frac{ 49 }{ 4 }\]Can you solve this for x? If not, I will gladly assist!!

Answer 10

x=47/4

Answer 11

It's actually\[x+\frac{ 1 }{ 2 }=\sqrt{\frac{ 49 }{ 4 }}\]See that?

Answer 12

Both 49 and 4 are perfect squares.

Answer 13

oh so it's x=3

Answer 14

Yes, it most certainly is! Good job!

Answer 15

is that the only answer or is this equation supposed to have two?

Answer 16

Thank You.

Answer 17

Actually it would have two. My bad.

Answer 18

Here they are:\[x+\frac{ 1 }{ 2 }=\pm \frac{ 7 }{ 2 }\]\[x=\frac{ 7 }{ 2 }-\frac{ 1 }{ 2 }=\frac{ 6 }{ 2 }=3\]OR\[x=-\frac{ 7 }{ 2 }-\frac{ 1 }{ 2 }=-\frac{ 8 }{ 2 }=-4\]Sorry.