You will plug these numbers into the quadratic formula to solve. a = 1, b = 8, c = 17. [tex] x=\frac{-8+/-\sqrt{8^2-4(1)(17)}}{2} [/tex], That simplifies down to [tex] x=\frac{-8+/-\sqrt{64-68}}{2} [/tex], and [tex] x=\frac{-8+/-\sqrt{-4}}{2} [/tex]. We say our roots are complex because we have a negative under the square root sign which we know is illegal. Unless we use the imaginary "i" and offset that negative. -1 is equal to i-squared, so let's make that replacement. [tex] x=\frac{-8+/-\sqrt{4*i^2}}{2} [/tex]. Both 4 and i-squared are perfect squares that can be pulled out as a 2 and an i, respectively. [tex] x=\frac{-8+/-2i}{2} [/tex]. Everything in the numerator can be reduced by the 2 in the denominator to give us these 2 solutions for x: [tex] x=-4+i [/tex] and [tex] x=-4-i [/tex].
Answer:
[x+(4-i)][x+(4+i)]
Step-by-step explanation:
Complete the square then regroup the first two terms. Add and subtract: (b/2)^2=(8/2)^2=16
x^2+8x+17=(x^2+8x)+17
=(x^2+8x+16)+17-16
=(x+4)^2+1 create a difference of squares using i^2= -1
=(x+4)^2 - (-1)
=(x+4)^2 - i^2 use the difference of two squares identity
=[(x+4) - i][(x+4)+i]
=[x+(4-i)][x+(4+i)]
