1. Solve
    2x2 - 6x + 8 = 0

    (1/2)(2x2 - 6x + 8) = (1/2)0

    x2 - 3x + 4 = 0

    (x2 - 3x + (-3/2)2) - (-3/2)2 + 4 = 0

    (x - 3/2)2 - 9/4 + 4 = 0

    (x - 3/2)2 + 7/4 = 0

    (x - 3/2)2 = -7/4

Now since every real number squared is greater than or equal to 0 this means that (x - 3/2)2 >= 0 and hence cannot be -7/4. Therefore, there are no real roots for that polynomial.
2. Solve
    x2 + 6x + 9 = 0

    (x2 + 6x + (6/2)2) - (6/2)2 + 9 = 0

    (x + 3)2 - 9 + 9 = 0

    (x + 3)2 = 0

    x + 3 = 0

    x = -3

The polynomial only has one root, which is when x = -3.

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