x = ±√( (4-8p)/4)
square it
x^2 = (4-8p)/4
x^2 = 1 - 2p
x^2 + 2p = 1
x^2 + 2p + p^2 = 1+p^2 , I completed the square
(x+p)^2 = 1+p^2
x+p = ±√(1+p^2)
x = -p ± √(1+p^2)
since 1+p^2 is always positive , x will always be real
to have equal roots
-p + √(1+p^2) = -p - √(1+p^2)
2√(1+p^2) = 0
1+p^2 = 0
p^2 = -1
which is not possible, thus no equal roots
we could have seen that from x^2 + 2p - 1 = 0
for equal roots b^2 - 4ac = 0
4p^2 - 4(1)(-1) = 0
4p^2 = -4
p^2 = -1 , as above
The solution to a quadratic equation is x is equals to 3 plus or minus square root of 4 minus 8p all over 4 where p is the element of Q.
Determine the value(s) of p that the roots of the equation are equal and non-real
1 answer