I put one complex solution and one real solution but was wrong?
What kind of solutions does the quadratic equation have.
- ComoLv 78 months ago
Assuming that should be shown as :-
x² - 4x + 10 = 0
x = [ 4 ± √(16 - 40) ] / 2
x = [ 4 ± j √24 ] / 2
x = 2 ± j √6
- Φ² = Φ+1Lv 78 months ago
Sorry Jeanelle, but you picked the only answer it couldn't be, given the three coefficients are real.
The line of reflection is x = -(-4)/(2×1) = 2, so the minimum value of this quadratic is 2²-4×2+10 = 6, and [x²] is positive, so x²-4x+10=0 has no real solutions. Both solutions are complex (and complex conjugates).
- DavidLv 78 months ago
If you mean x^2 -4x +10 = 0 then it has no real solutions because its discriminant of b^2 -4ac = -24 which is less than 0 meaning it has no real roots.
Discriminant less than 0 no real roots
Discriminant equal to 0 equal roots
Discriminant more than 0 two different roots
- Ian HLv 78 months ago
Just use the quadratic formula. You will get the pair of complex conjugates
2 +√(6)i and 2 -√(6)I
With a quadratic you either get 2 reals or 2 complex conjugates
- What do you think of the answers? You can sign in to give your opinion on the answer.
- llafferLv 78 months ago
Complex (and irrational) roots of polynomial equations always come in conjugate pairs.
So in a 2nd degree polynomial, you can't just have 1 complex and 1 real root. They are either both complex or both real (or the case where both real roots are the same value resulting in a single real root).
To determine this for a quadratic, look at its discriminant:
b² - 4ac
If this is > 0, you have two real roots.
If this is = 0, you have one unique real root.
If this is < 0, you have 0 real roots (both are complex)
a = 1, b = -4, c = 10, we get:
b² - 4ac
(-4)² - 4(1)(10)
16 - 40
Since this is negative, there are no real roots. Both are complex.