Answer:
t=-3
Step-by-step explanation:
2x²-4x-t=0
disc=b²-4ac=(-4)²-4×2×(-t)=16+8t
it has no solution if 16+8t<0
8t<-16
t<-2
so t=-3
2.
[tex]\sqrt{x-a} =x-4\\if a=2\\\sqrt{x-2} =x-4\\squaring\\x-2=x^2-8x+16\\x^2-9x+18=0\\[/tex]
x²-6x-3x+18=0
x(x-6)-3(x-6)=0
(x-6)(x-3)=0
x=6,3
when x=6
√(6-2)=6-4
√4=2
2=2
when x=3
√(3-2)=3-4
√1=-1
1=-1
which is not true.
Hence x=3 is an extraneous solution.
x=6 is real solution.
