Question

Consider the quadratic equation. x^2=4x-5. How many solutions does the equation have?

a.) one real solution
b.) two real solutions c.)no real solutions
d.)cannot be determined

Answer 1

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=
`(-b+ √(b^2-4ac) )/(2a)` or
`(-b- √(b^2-4ac) )/(2a)`

si ax^2+bx+c
so a=1
b=-4
c=5
input

`(-(-4)+ √(-4^2-4(1)(5)) )/(2(1))`=
`(4+ √(16-20) )/(2(1))`=
`(4+ √(-4) )/(2)`=
`(4+ √(4) times √(-1) )/(2)`
`(4+2 times √(-1) )/(2)= (6 times √(-1) )/(2)=3 times √(-1) [\tex][\tex]√(-1)` representeds by 'i' so solution is 3i

then if other way around then wyou would do

`(-(-4)- √(-4^2-4(1)(5)) )/(2(1))`=
`(4- √(16-20) )/(2(1))= (4- √(-4) )/(2) =(4- √(4) times √(-1) )/(2)= (4-2 times √(-1) )/(2)=(2 √(-1) )/(2)= √(-1)` and [\tex]\sqrt{-1} [/tex] is represented by i


the solution is x=3i or i (i=
`√(-1)`)
but i is not real, it is imaginary so there are no real solution so the answer is C