Question

If the squared difference of the zeros of the quadratic polynomial f(x) = x² + px + 45 is equal to 144, find the value of p.

Answer 1

Given α and β are the zeros of the quadratic polynomial f(x) = x² + px + 45

`alpha+beta=-"coefficient of x"/("coefficient of "x^2)`

`=(-p)/1`

= -p

`alphabeta="constant term"/("coefficient of "x^2)`

`=45/1`

= 45

we have,

`(alpha-beta)^2=alpha^2+beta^2-2alphabeta`

`144=(alpha+beta)^2-2alphabeta-2alphabeta`

`144=(alpha+beta)^2-4alphabeta`

Substituting `alpha+beta=-p " and "alphabeta=45`

then we get,

`144=(-p)^2-4xx4`

`144=p^2-4xx45`

`144=p^2-180`

`144+180=p^2`

`324=p^2`

`sqrt(18xx18)=pxxp`

`+-18=p`

Hence, the value of p is ±18.