Question

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

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=$\frac{-b+ \sqrt{b^2-4ac} }{2a}$ or $\frac{-b- \sqrt{b^2-4ac} }{2a}$

si ax^2+bx+c
so a=1
b=-4
c=5
input
$\frac{-(-4)+ \sqrt{-4^2-4(1)(5)} }{2(1)}$=$\frac{4+ \sqrt{16-20} }{2(1)}$=$\frac{4+ \sqrt{-4} }{2}$=$\frac{4+ \sqrt{4} times \sqrt{-1} }{2}$ $\frac{4+2 times \sqrt{-1} }{2}= \frac{6 times \sqrt{-1} }{2}=3 times \sqrt{-1} [\tex][\tex]\sqrt{-1}$ representeds by 'i' so solution is 3i

then if other way around then wyou would do
$\frac{-(-4)- \sqrt{-4^2-4(1)(5)} }{2(1)}$=$\frac{4- \sqrt{16-20} }{2(1)}= \frac{4- \sqrt{-4} }{2} =\frac{4- \sqrt{4} times \sqrt{-1} }{2}= \frac{4-2 times \sqrt{-1} }{2}=\frac{2 \sqrt{-1} }{2}= \sqrt{-1}$ and [\tex]\sqrt{-1} [/tex] is represented by i


the solution is x=3i or i (i=$\sqrt{-1}$)
but i is not real, it is imaginary so there are no real solution so the answer is C