Determine the number of solutions of x^2 - 5x + 40 = 0
answers could be
0
1(double root)
2
I'm not sure on this one where to start
I know I think I set it up quadratic formula but then I'm baffled
the √(b^2 - 4ac) part of the formula is called the
"discriminant"
If its value > 0 you will have 2 different real roots
if its value = 0 you will one real root (a double or 2 equal roots)
if its value < 0 there will be two different imaginary or complex roots, or no real roots.
so evaluate (-5)^2 - 4(1)(40) and state your answer
Is this what you are looking for?
The sign of the discriminant determines the number and type of solutions.
If +, there are two real solutions
If 0, there is one real solution
If -, there are two imaginary solutions
I solved this and I came up with two nonreal solutions, correct so the answer to my problem above would be then be "0" soltions correct?
there are no real solutions, but there are 2 imaginary solutions
b^2 - 4ac < 0,
I got as my solutions
x = 1/2(5 +- 3i sqrt(15))
But this is where I get confused. If there are two imaginary solutions, then do I say I have 2, 1 or 0 solutions-that's what I'm not getting, still? in regards to the original problem which is statd as a multiple choice above
The answer would be 2.
If the question had been worded how many
'real solutions' (which it isn't) your answer would be 0, since the answer is an 'imaginary number' not 'real number'.
Do you understand the difference between real number (solutions) and imaginary number (solutions)?
I am not a tutor, so maybe
Reiny will answer and can explain more clearly.
Above I stated I had worked this problem out and there were two IMAGINARY solutions but I'm still confused about the question which asks how many solutions arethere to the problem-should I put "0" or do the two imaginary solutions count and I should put "2". That's what I'm still confused on
Thank you
Also, I think I confused you above by saying "there are no real solutions, but there are 2 imaginary solution".
I should have just said 2 imaginary solutions, since that was what you were asking.
sorry to confuse you.
yes, the imaginary count as solutions, so the answer is 2.
Thank you, that was the part that after I solved itand came up with the two imaginary, was still confusing me-
Thanks again