Question

find the natural of the root of quadrilateral equation X square - 6 X + 9=0​

Answer 1

Answer:

roots=3,3. 3*3=9, -3-3= -6.so the roots of x ^2-6x+9 is 3, 3.

Answer 2

Answer:

x2-6x+9=0

comparing with ax2+bx+c=0 we get,

a=1,b= -6,c=9

Now,

b^2-4ac=(-6)^2-4(1)(9)

=36-36

=0

since b^2-4ac=0

therefore,

roots are real and equal