# I put one complex solution and one real solution but was wrong?

What kind of solutions does the quadratic equation have.

x^{^2}-4x+10=0

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Assuming that should be shown as :-

x² - 4x + 10 = 0

x = [ 4 ± √(16 - 40) ] / 2

x = [ 4 ± j √24 ] / 2

x = 2 ± j √6

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• 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).

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• 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

• It would be easier to say that x^2 - 4x + 10 is always positive, greater than zero, for any real x.

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• 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

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• 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)

Using:

a = 1, b = -4, c = 10, we get:

b² - 4ac

(-4)² - 4(1)(10)

16 - 40

-24

Since this is negative, there are no real roots.  Both are complex.

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