Question

(х-4)^4-8(х-4)^2-9=0

Answer 1

(Х-4)^4-8(х-4)^2-9=0

(Х-4)^2=t

t^2-8t-9=0

D=64+36=100

t1=(8+10)/2=18/2=9

t2=(8-10)/2=-2/2=-1 неудов.

(Х-4)^2=9

Х-4=3

Х=3+4

Х=7

(Х-4)=-3

Х=-3+4

Х=1

Answer 2

To solve the equation (х-4)^4 - 8(х-4)^2 - 9 = 0, we can use a substitution to simplify the equation and make it easier to solve.

Let's make a substitution: Let y = (х-4)^2.

Now, our equation becomes y^2 - 8y - 9 = 0.

We can solve this quadratic equation for y using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a,

where a = 1, b = -8, and c = -9.

y = (8 ± √((-8)^2 - 4 * 1 * (-9))) / 2 * 1 y = (8 ± √(64 + 36)) / 2 y = (8 ± √100) / 2 y = (8 ± 10) / 2

Now we have two possible values for y:

1. y = (8 + 10) / 2 = 18 / 2 = 9
2. y = (8 - 10) / 2 = -2 / 2 = -1

Recall that y = (х-4)^2. Now, we can use these values of y to find the values of х:

1. y = (х-4)^2 = 9 Taking the square root of both sides: √((х-4)^2) = √9 х - 4 = ±3 Two possible solutions: х = 4 + 3 = 7 х = 4 - 3 = 1

2. y = (х-4)^2 = -1 Since a square cannot be negative, there are no real solutions in this case.

So, the solutions to the equation are х = 7 and х = 1.