What is the quadratic equation that has a leading coefficient of 1 and solutions of $3+sqrt2$ and $3-sqrt2$ ?

Question

1237 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-16T08:24:38

$x^2+6x+7=0$

$"if "alpha " & "beta " are the roots of a quadratic then its equation can be written as"$

$x^2-(alpha+beta)x+alphabeta=0$

see proof below
we ahve roots as being

$(3+sqrt2),(3-sqrt2)$

the quadratic is then

$x^2-(3+sqrt2+3-sqrt2)x+(3+sqrt2)(3-sqrt2)$ =0

$x^2-6x+(9-2)=0$

$x^2-6x+7=0$

proof

$alpha, beta " roots of a quadratic"$

$=>(x-alpha)(x-beta)=0$

$x^2-betax-alphax+alphabeta=0$

$=>x^2-(alpha+beta)x+alphabeta=0$

What is the quadratic equation that has a leading coefficient of 1 and solutions of 3+sqrt23+sqrt2 and 3-sqrt23-sqrt2?