Question 839475
Given that the quadratic equation x^2-2x-5=0 has 2 different roots, and a second quadratic equation has 2 roots, each of which 2 less than the corresponding root of the given quadratic equation. If the second quadratic equation is x^2+ax+b=0, find the value of a and b.


{{{x^2-2x-5=0}}}

{{{x^2-2x+1-1-5=0}}}


{{{(x-1)^2-6=0}}}

{{{(x-1)^2=6}}}

(x-1) = +/- sqrt(6)



The roots of the second equation will be 

{{{sqrt(6)-2}}}, {{{-sqrt(6)-2}}}

sum  of roots = -4

product of the roots = {{{sqrt(6)-2}}}*{{{-sqrt(6)-2}}}

=-2

The general equation is
x^2-sum of roots(x)+product of the roots=0

{{{x^2-(-4)x+(-2)=0}}}

{{{x^2+4x-2=0}}}

comparing the equation with

x^2+ax+b=0

we get
a=4, b=-2