Question

How to create a quadratic equation when given is that one of the solutions must be 1/(3+2sqrt2) . Show steps.

Answer 1

hw many times will u ask the same question?stop spamming!

Answer 2

Spamming? I didn't get an answer last time.

Answer 3

so wait u will get an answer..dont post the same question again

Answer 4

y = (x- 1/(3+2sqrt2))(x-a)
a can be anything

Answer 5

Thanks!

Answer 6

you could also rationalize it
\[\frac{1}{3+2\sqrt{2}}*\frac{3-2\sqrt{2}}{3-2\sqrt{2}} = 3-2\sqrt{2}\]
so then
\[x = 3\pm 2\sqrt{2}\]
\[x-3 = \pm2\sqrt{2}\]
\[(x-3)^{2} = 8\]
\[x^{2} -6x+1 = 0\]

Answer 7

Thank you!