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

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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#