Problem: The values of k for which the equation 2x2βkx+x+8=0 will have real and equal roots are:
Answer Choices:
A. 9 and β7
B. only β7
C. 9 and 7
D. β9 and β7
E. only 9
Solution:
2x2βx(kβ1)+8=0. For real, equal roots, the discriminant (kβ1)2β64=0 and kβ1=Β±8.β΄k=9 or β7.