find the zeros:
x^2 - 2x + 8
Pairs of integers that multiply to give +8:
1, 8
-1, -8
2, 4
-2, -4
Hmm....none of these add to give -2, so looks like this expression is unfactorable.
Plan B: Use the quadratic formula.
-b +-sq root of b^2-4ac all of this divided by 2a
2 +- sq root (-2)^2 -4(1)(8)
all divided by 2(1)
2 +- sq root 4-32
2+- sq root -28 all divided by 2
2+- 2i sq root of 7 comes from
4(-1)(7)
all divided by 2 reduces to
1 +i sq rt 7, 1-i sq rt 7 are the roots. or are the values of x.
Excellent!
To find the zeros of the quadratic equation x^2 - 2x + 8, you need to solve for x when the equation is equal to zero.
Step 1: Write the equation: x^2 - 2x + 8 = 0
Step 2: Use the quadratic formula to solve for x, which is -b ± √(b^2 - 4ac) / 2a. In this case, a = 1, b = -2, and c = 8.
x = (-(-2) ± √((-2)^2 - 4(1)(8))) / (2(1))
Simplifying this, we get:
x = (2 ± √(4 - 32)) / 2
x = (2 ± √(-28)) / 2
Step 3: Since we have a negative value inside the square root (√(-28)), it means that this quadratic equation has no real roots. The zeros are imaginary or complex numbers. So, there are no real solutions or zeros for this quadratic equation.