Question

Sqrt(x^3+8)+(x^3+8)^1/4=6

Answer 1

Ответ: x=2.

Объяснение:

√(x³+8)+⁴√(x³+8)=6   ОДЗ: x³+8≥0    x³≥-8       x³≥-2³       x≥-2.

Пусть ⁴√(x³+8)=t≥0      ⇒

t²+t-6=0      D=25      √D=5

t₁=⁴√(x³+8)=2      (⁴√(x³+8))⁴=2⁴        x³+8=16     x³=8        x=2.

t₂=⁴√(x³+8)=-3  ∉.

Answer 2

To solve the equation: √(x^3 + 8) + (x^3 + 8)^(1/4) = 6, we can follow these steps:

Step 1: Substitute (x^3 + 8) with a variable, let's say u. Step 2: Rewrite the equation with the new variable u. Step 3: Solve the new equation for u. Step 4: Substitute back u with (x^3 + 8) to find the solutions for x.

Let's go through the steps in detail:

Step 1: Substitute (x^3 + 8) with u:

u = x^3 + 8

Step 2: Rewrite the equation with the new variable u:

√u + u^(1/4) = 6

Step 3: Solve the new equation for u:

Let's isolate u on one side of the equation:

√u = 6 - u^(1/4)

Now, square both sides to eliminate the square root:

u = (6 - u^(1/4))^2

Expand the right side:

u = 36 - 12u^(1/4) + u^(1/2)

Now, let's make another substitution to simplify the equation. Let's say v = u^(1/4). Then we have:

u^(1/2) = v^2

Now, the equation becomes:

v^4 = 36 - 12v + v^2

Move all terms to one side of the equation:

v^4 - v^2 + 12v - 36 = 0

Step 4: Solve the equation for v:

This is a quadratic equation in terms of v^2. Let's factor it:

(v^2 - 6)(v^2 + 6) = 0

Now, solve for v^2:

1. v^2 - 6 = 0 v^2 = 6 v = ±√6

2. v^2 + 6 = 0 v^2 = -6 (Since v represents the fourth root of u, v^2 can't be negative) This solution is not valid.

Now, remember v = u^(1/4). So, we have:

1. u^(1/4) = √6 u = (√6)^4 u = 6^2 u = 36

Now, revert the substitution u = x^3 + 8:

x^3 + 8 = 36

Subtract 8 from both sides:

x^3 = 36 - 8 x^3 = 28

Finally, take the cube root of both sides to get the solutions for x:

x = ∛28 ≈ 3.036

So, the solutions for the equation are approximately x = 3.036 and two complex solutions (which we discarded in this real-number domain analysis).