Question

ILL GIVE BRAINLIST PLS HELP

Answer 1

Answer:

Last option) 3

Step-by-step explanation:

For having no real solutions, discriminant of this equation must be be less than 0.

D=$\sqrt{b^{2}-4ac }$

b²-4ac<0

b²<4ac

b²/4a<c

c>b²/4a

c>$\frac{(-4^{2} )}{8}$

c>16/8

c>2

Answer 2

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.

$\sqrt{x-a} =x-4\\if a=2\\\sqrt{x-2} =x-4\\squaring\\x-2=x^2-8x+16\\x^2-9x+18=0$

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.