Question

It is possible for a quadratic equation to have no real-number solutions.

Solve.

u ^ 2 + 2u - 4 = 0

Answer 1

\[u ^ 2 + 2u - 4 = 0\]\[Δ=2^2-4\times(-4)=20\]\[x_1,_2=(-2\pm \sqrt{20})/2\times1\]
\[x=-1+\sqrt{5}\]or\[x=-1-\sqrt{5}\]

Answer 2

It is possible for the quadratic equation to have no real solution, if b^2-4ac < 0 then you will get an imaginary solution. (This also means the graph does not cut the x-axis, and i above or below it)

Answer 3

u can use quadratic formula here:

u =[ -2 +- sqrt(2^2 - 4 *1 * -4) ] / 2
= -1 +- sqrt 5

Answer 4

thankz jimmy !