Question

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

Answer 1

If one solution is 1/(3+2sqrt2 then there must be another root of 1/3-2sqrt2 since they always come in pairs

Answer 2

yeah mertsj is rite..but mertsj when roots are real they are not in pairs

Answer 3

Remember the quadratic formula
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 4

See that plus or minus thing? That's why they always come in pairs.

Answer 5

i think we should rationalize first

Answer 6

Can you show all the steps?

Answer 7

@king If roots are irrational they are in pairs.

Answer 8

yeah exactly what i was sayin!

Answer 9

\[\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}\]

Answer 10

Alright, no idea.

Answer 11

@king It's not what you were saying. You said if the roots are real, they are not in pairs and irrational numbers are real numbers.

Answer 12

so Erko. One of the roots is 3-2sqrt2. That means the other one is 3+2sqrt2

Answer 13

@Mertsj sry i meant that only!

Answer 14

So the equation is x minus the first root times x - the second root

Answer 15

\[[x-(3-2\sqrt{2})][x-(3+2\sqrt{2})]=0\]

Answer 16

and there is your equation.