Quadratic eqn qstn about roots?

Question

Given that k is real, fint the set of values of k for which the roots of the quadratic equation (1+2k)x²-10x+(k-2)=0

a)have a sum which is greater than 5

(pls show working so i could understand)

Answer 1

apply the quadratic formula to get 2 answers in terms of k

Add THE TWO answers together and solve sum > 5

Answer 2

(1+2k)x²-10x+(k-2)=0

ax^2+bx+c =0

x1+x2= -b/a

b=10; a=1+2k;

-b/a>5

10/(1+2k)>5

divide both sides by 5

2/(1+2k)>1

2/(1+2k) -1>0

(2-1-2k)/(1+2k) >0

(1-2k)/(1+2k) >0

N>0 => 1-2k>0; -2k>-1; 2k<1; k<1/2;

D>0 => 1+2k>0; k>-1/2;

____ -1/2________1/2_______ k

__+__ I____+_____ I _ _ _ - _ _ N>0

_ _-_ _I____+_____ I ____+___ D>0

-1/2<k<1/2;