Question

how can i solve x^2-3x+1 using "complete the square" method?

Answer 1

First, isolate the x-terms. so subtract one from both sides.
Next, complete the sq. by dividing the coefficient on x (bx in the formula ax+bx+cx) by 2 and then squaring that number. whatever you come up with needs to be added to both sides.

For example, y -3=x^2+4x
the 4x turns to 2x when I divide it, and then that sqared is 4, so I add 4 to both sides
y-3+4=x^2+2x+4

Now, simplify the equation.

Answer 2

Does that make some sense? (:

Answer 3

gonna read it a couple of times, lol

Answer 4

okay. khanacademy.com has some great lessons on "completing the square" too @sta880

Answer 5

so far i've got something like this x^2-3x+2.25=-1+2.25 , is this the right path?

Answer 6

I would think so. Are you on FLVS?

Answer 7

x-1.25=sqrt1.25, answer must be in significant figures, so should I just mark it as unsolveable?

Answer 8

I'm not sure. @ranga can you help us?

Answer 9

does this mean "solve for \(x\): \(x^2-3x+1 =0\)?

Answer 10

oh, yeah, sorry

Answer 11

if you want the complete the square method, first step is subtract \(1\) from both sides and write
\[x^2-3x=-1\] then since half of \(3\) is \(\frac{3}{2}\) go right to
\[(x-\frac{3}{2})^2=-1+(\frac{3}{2})^2=-1+\frac{9}{4}=\frac{5}{4}\]

Answer 12

once you have
\[(x-\frac{3}{2})^2=\frac{5}{4}\] take the square root of both sides, don't forget the \(\pm\) and write
\[x-\frac{3}{2}=\frac{\sqrt5}{2}\]

Answer 13

oops i forgot the \(\pm\) it should be
\[x-\frac{3}{2}=\pm\frac{\sqrt5}{2}\] then finish by adding \(\frac{3}{2}\) to both sides getting
\[x=\frac{3}{2}\pm\frac{\sqrt5}{2}\] or if you prefer
\[x=\frac{3\pm\sqrt5}{2}\]