Question

find the value for c that makes the equation have
a. two real solution
b.one real solution
c. two imaginary solution
x^2-6x+c=0

Answer 1

when c=9

x²-6x+9=0

factorizing,

x²-3x-3x+9=0

x(x-3)-3(x-3)=0

(x-3)(x-3)=0

the zeros are 3 and 3.so the zeros are equal

when c=5

x²-6x+5=0

factorizing,

x²-1x-5x+5=0

x(x-1)-5(x-1)=0

(x-1)(x-5)=0

x=1 and 5

the zeros are real numbers

when x=10

x²-6x+10=0

since factorization is noy possible

use the equation

x=[-b±√(b²-4ac)]/2a

x=[-(-6)±√(-6)²-4×1×10)]/2

x=(6±√(36-40))/2

x=(6±√-4)/2

x=3±2i

x= 3+2i or 3-2i

pls mark brainliest