SOLUTION: 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 quadr

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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.
Answer by mananth(16946) About Me  (Show Source):
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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%5E2-2x-5=0
x%5E2-2x%2B1-1-5=0

%28x-1%29%5E2-6=0
%28x-1%29%5E2=6
(x-1) = +/- sqrt(6)


The roots of the second equation will be
sqrt%286%29-2, -sqrt%286%29-2
sum of roots = -4
product of the roots = sqrt%286%29-2*-sqrt%286%29-2
=-2
The general equation is
x^2-sum of roots(x)+product of the roots=0
x%5E2-%28-4%29x%2B%28-2%29=0
x%5E2%2B4x-2=0
comparing the equation with
x^2+ax+b=0
we get
a=4, b=-2