What is the value of #E# so that # 4x^2-4x+E = 0 # has distinct real roots?

Question

830 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-05T00:57:02

# E lt 1 #

We have a quadratic:

# 4x^2-4x+E = 0 #

We have two unique (real) solutions if and only if the discriminant of the quadratic is positive. i.e

# Delta = b^2-4ac gt 0 #
# => (-4)^2-4(4)(E) gt 0 #
# :. 16-16E gt 0 #
# :. 1-E gt 0 #
# :. E lt 1 #

We can confirm this graphically, as #E=1# should present two coincidental solutions, and #E gt 1 # no solutions:

Case #E=1# - One Repeated Solution
graph{4x^2-4x+1 [-3, 3, -2, 5]}

Case #E=1.1# - No Solutions
graph{4x^2-4x+1.1 [-3, 3, -2, 5]}

Case #E=0.9# - Two Solutions
graph{4x^2-4x+0.9 [-3, 3, -2, 5]}