Problem: Given the equations x2+kx+6=0 and x2−kx+6=0. If, when the roots of the equations are suitably listed, each root of the second-equation is 5 more than the corresponding root of the first equation, then k equals:
Answer Choices:
A. 5
B. −5
C. 7
D. −7
E. none of these
Solution:
Let the roots of the first equation be r and s. Then r+s=−k and rs=6. Then, for the second equation, r+5+s+5=k and (r+5)(s+5)=6.∴rs+5(r+s)+25=6,6+5(−k)+25=6,k=5.
or
Let r be one of the roots of the first equation; then r+5 is the associated root of the second equation. ∴(r+5)2−k(r+5)+6=0,r2+(10−k)r+31−5k=0. But r satisfies the first equation so that r2+kr+6=0.∴10−k=k and 31−5k=6. Each of these equations leads to the result k=5.