## anonymous 4 years ago Find a number k such that the given equation has exactly one real solution. x^2-kx+4=0

1. anonymous

the answer is 4 or -4

2. anonymous

you just use the discriminant

3. campbell_st

use b^2 - 4ac = 0 from the GQF..... or (-k)^2 - 4x1x4 = 0 and solve

4. ash2326

ax^2+bx+c=0 for one real solution D=b^2-4ac=0 here a=1 b=-k c=4 K^2-4*1*4=0 k^2-16=0 so $$k=\pm 4$$