x^2=4x-5
subtract 4x from both sides
x^2-4x=-5
add 5 to both sides
x^2-4x+5=0
input into quadratic formula which is x=
or
si ax^2+bx+c
so a=1
b=-4
c=5
input
=
=
=
![\frac{4+2 times \sqrt{-1} }{2}= \frac{6 times \sqrt{-1} }{2}=3 times \sqrt{-1} [\tex][\tex]\sqrt{-1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%2B2%20times%20%20%5Csqrt%7B-1%7D%20%20%7D%7B2%7D%3D%20%20%5Cfrac%7B6%20times%20%20%5Csqrt%7B-1%7D%20%20%7D%7B2%7D%3D3%20times%20%20%5Csqrt%7B-1%7D%20%5B%5Ctex%5D%5B%5Ctex%5D%5Csqrt%7B-1%7D%20%20)
representeds by 'i' so solution is 3i
then if other way around then wyou would do
=
and [\tex]\sqrt{-1} [/tex] is represented by i
the solution is x=3i or i (i=
)
but i is not real, it is imaginary so there are no real solution so the answer is C
Since the roots are 2,3i, and -3i, you can write it as
(x-2)(x-3i)(x+3i)=(x-2)(x^2+9)=x^3-2x^2+9x-18.
Answer:
Step-by-step explanation:
FIRST multiply the coefficients together:
15a^2*b^7*a^3*b^8
Add up the exponents of the "a" terms:
15*a^(2 + 3), and, finally, add up the exponents of the "b" terms:
15*a^5*b^15
