Question

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

Answer 1

`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`

Answer 2

`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`