How do you find a quadratic polynomial with integer coefficients which has #x=3/5+-sqrt29/5# as its real zeros?

2 Answers
Sep 21, 2017

Answer:

#5x^2-6x-4#

Explanation:

#"given the zeros ( roots) of a quadratic "x=a,x=b#

#"then the factors are "(x-a),(x-b)#

#"and the quadratic is the product of the factors"#

#rArrp(x)=(x-a)(x-b)#

#"here the roots are "x=3/5+sqrt29/5" and "x=3/5-sqrt29/5#

#rArr"factors "(x-(3/5+sqrt29/5)),(x-(3/5-sqrt29/5))#

#rArrp(x)=(x-3/5-sqrt29/5)(x-3/5+sqrt29/5)#

#color(white)(rArrp(x))=(x-3/5)^2-(sqrt29/5)^2#

#color(white)(rArrp(x))=x^2-6/5x+9/25-29/25#

#color(white)(rArrp(x))=x^2-6/5x-4/5to(xx5)#

#color(white)(rArrp(x))=5x^2-6x-4#

Sep 21, 2017

Answer:

#5x^2-6x-4=0#

Explanation:

if #alpha " & "beta" are the roots of a quadratic equation"#

the equation can be written as

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

we have

#alpha=3/5+sqrt29/5#

#beta=3/5-sqrt29/5#

#alpha+beta=3/5+cancel(sqrt29/5)+3/5-cancel(sqrt29/5)#

#alpha+beta=6/5#

#alphabeta=(3/5+sqrt29/5)(3/5-sqrt29/5)#

#=9/25-29/25=-20/25=-4/5#

#:.x^2-(alpha+beta)x+alphabeta#

becomes

#x^2-6/5x-4/5=0#

#=>5x^2-6x-4=0#