Question

section 8-4 exercises
solve.

1. \\((5t + 7)^2 + 4(5t + 7) - 5 = 0\\)
2. \\(x^{2/3} - 4x^{1/3} - 21 = 0\\)
3. \\(x^4 + 5x^2 - 36 = 0\\)
4. \\(x^{2/3} + 6x^{1/3} - 16 = 0\\)

tough one. use factoring and the quad

1. \\(x^6 - 7x^3 - 8 = 0\\)

Explanation:

Let's solve problem 9: \( x^6 - 7x^3 - 8 = 0 \) using substitution.

Step 1: Substitution

Let \( u = x^3 \). Then \( x^6 = (x^3)^2 = u^2 \). Substitute these into the equation:

$$ u^2 - 7u - 8 = 0 $$

Step 2: Factor the quadratic equation

We need to factor \( u^2 - 7u - 8 \). We look for two numbers that multiply to -8 and add to -7. The numbers are -8 and 1. So,

$$ u^2 - 7u - 8 = (u - 8)(u + 1) = 0 $$

Step 3: Solve for \( u \)

Set each factor equal to zero:

• \( u - 8 = 0 \) gives \( u = 8 \)
• \( u + 1 = 0 \) gives \( u = -1 \)

Step 4: Substitute back \( u = x^3 \) and solve for \( x \)

• For \( u = 8 \): \( x^3 = 8 \). Taking the cube root of both sides, \( x = \sqrt[3]{8} = 2 \).
• For \( u = -1 \): \( x^3 = -1 \). Taking the cube root of both sides, \( x = \sqrt[3]{-1} = -1 \).

(Note: In the real number system, the cube root of a negative number is negative, so these are the only real solutions. If we consider complex numbers, there would be more solutions, but typically for such problems, real solutions are expected unless stated otherwise.)

Final Answer

The solutions are \( \boldsymbol{x = 2} \) and \( \boldsymbol{x = -1} \).

Answer:

Let's solve problem 9: \( x^6 - 7x^3 - 8 = 0 \) using substitution.

Step 1: Substitution

Let \( u = x^3 \). Then \( x^6 = (x^3)^2 = u^2 \). Substitute these into the equation:

$$ u^2 - 7u - 8 = 0 $$

Step 2: Factor the quadratic equation

We need to factor \( u^2 - 7u - 8 \). We look for two numbers that multiply to -8 and add to -7. The numbers are -8 and 1. So,

$$ u^2 - 7u - 8 = (u - 8)(u + 1) = 0 $$

Step 3: Solve for \( u \)

Set each factor equal to zero:

• \( u - 8 = 0 \) gives \( u = 8 \)
• \( u + 1 = 0 \) gives \( u = -1 \)

Step 4: Substitute back \( u = x^3 \) and solve for \( x \)

• For \( u = 8 \): \( x^3 = 8 \). Taking the cube root of both sides, \( x = \sqrt[3]{8} = 2 \).
• For \( u = -1 \): \( x^3 = -1 \). Taking the cube root of both sides, \( x = \sqrt[3]{-1} = -1 \).

(Note: In the real number system, the cube root of a negative number is negative, so these are the only real solutions. If we consider complex numbers, there would be more solutions, but typically for such problems, real solutions are expected unless stated otherwise.)

Final Answer

The solutions are \( \boldsymbol{x = 2} \) and \( \boldsymbol{x = -1} \).