What is the value of #(alpha - beta)# ?

If the roots of the equation #x^2-22x+105=0# are #alpha , beta# then what is the value of #(alpha - beta)#

2 Answers
Mar 20, 2018

#alpha-beta=8#

Explanation:

For the equation #x^2+lx+m=0#

sum of roots is #-l# and product of roots is #m#.

Therefore, as for #x^2-22x+105=0# roots are #alpha# and #beta#

hence #alpha+beta=-(-22)=22# and #alphabeta=105#

As #(alpha+beta)^2=(alpha-beta)^2+4alphabeta#

#22^2=(alpha-beta)^2+4*105#

or #(alpha-beta)^2=22^2-420=484-420=64#

and #alpha-beta=8#

One could say that we can also have #alpha-beta=-8#, but observe that #alpha# and #beta# are not in any particular order. The roots of equation are #15# and#7# and their #alpha-beta# could be #15-7# as well as #7-15#, it deends on what you choose as #alpha# and #beta#.

Mar 20, 2018

If #(alpha>beta)#, then ,#(alpha-beta)=8#

Explanation:

If the quadratic equation #ax^2+bx+c=0#, has roots #alpha and beta, #then #alpha+beta=-b/a and alpha*beta=c/a.#
Here,

#x^2-22x+105=0=>a=1 , b=-22 , c=105#

So,
#alpha+beta=-(-22)/1=22 ,and alphabeta=105/1=105#
Now,
#(alpha-beta)=sqrt((alpha+beta)^2-4alphabeta#,...# where,(alpha>beta)#

#(alpha-beta)=sqrt((22)^2-4(105))#

#(alpha-beta)=sqrt(484-420)=sqrt64=8#