Solving the Equation \(8k^6 + 15k^3 – 2 = 0\)

We are asked to find the positive value of \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right) \) given the equation \( 8k^6 + 15k^3 – 2 = 0 \). This equation looks complicated because of the powers of \(k\), but we can simplify it using a substitution.

Using Substitution to Solve the Equation

Notice that the equation involves \(k^6\) and \(k^3\). We can write \(k^6\) as \((k^3)^2\). This suggests a substitution to turn the equation into a more familiar form, like a quadratic equation.

Let \( x = k^3 \). Substituting this into the given equation \( 8k^6 + 15k^3 – 2 = 0 \), we get:

\( 8(k^3)^2 + 15(k^3) – 2 = 0 \)

\( 8x^2 + 15x – 2 = 0 \)

This is a standard quadratic equation in terms of \(x\).

Solving the Quadratic Equation \(8x^2 + 15x – 2 = 0\)

We can solve this quadratic equation for \(x\) using factorization or the quadratic formula. Let's use factorization:

We need to find two numbers that multiply to \(8 \times (-2) = -16\) and add up to \(15\). These numbers are \(16\) and \(-1\).

Rewrite the middle term \(15x\) as \(16x - x\):

\( 8x^2 + 16x - x – 2 = 0 \)

Group the terms and factor:

\( (8x^2 + 16x) + (-x – 2) = 0 \)

Factor out common terms from each group:

\( 8x(x + 2) - 1(x + 2) = 0 \)

Factor out the common binomial term \((x + 2)\):

\( (8x - 1)(x + 2) = 0 \)

This equation gives us two possible values for \(x\):

• \( 8x - 1 = 0 \implies 8x = 1 \implies x = \frac{1}{8} \)
• \( x + 2 = 0 \implies x = -2 \)

Finding the Values of \(k\)

Now we substitute back \( x = k^3 \) to find the possible values of \(k\).

Case 1: \( x = \frac{1}{8} \)

\( k^3 = \frac{1}{8} \)

Taking the cube root of both sides:

\( k = \sqrt[3]{\frac{1}{8}} \)

\( k = \frac{1}{2} \)

Case 2: \( x = -2 \)

\( k^3 = -2 \)

Taking the cube root of both sides:

\( k = \sqrt[3]{-2} \)

Note that \(\sqrt[3]{-2}\) is a real number, approximately \(-1.26\).

Calculating the Value of \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right) \)

We need to find the positive value of \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right) \). Let's evaluate this expression for each value of \(k\) we found.

For \( k = \frac{1}{2} \):

\( k + \frac{1}{k} = \frac{1}{2} + \frac{1}{\frac{1}{2}} \)

\( = \frac{1}{2} + 2 \)

\( = 2.5 \)

In mixed fraction form, \( 2.5 = 2\frac{1}{2} \). This value is positive.

For \( k = \sqrt[3]{-2} \):

\( k + \frac{1}{k} = \sqrt[3]{-2} + \frac{1}{\sqrt[3]{-2}} \)

\( = \sqrt[3]{-2} + \sqrt[3]{-\frac{1}{2}} \)

Both \(\sqrt[3]{-2}\) and \(\sqrt[3]{-\frac{1}{2}}\) are negative numbers. Their sum will be a negative number. Since we are looking for the positive value of \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right) \), this case does not give the required answer.

Therefore, the positive value of \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right) \) is \( 2\frac{1}{2} \).

Let's verify this with the given options.

Our calculated positive value \( 2\frac{1}{2} \) matches Option 1.

Revision Table: Key Concepts Revisited

Additional Information: Roots of Equations and Real Values

When solving equations like \(k^3 = a\), there is always at least one real root. If \(a\) is positive, the real root \(\sqrt[3]{a}\) is positive. If \(a\) is negative, the real root \(\sqrt[3]{a}\) is negative. In our problem, \(k^3 = 1/8\) gives the positive real root \(k=1/2\), and \(k^3 = -2\) gives the negative real root \(k=\sqrt[3]{-2}\).

The question specifically asked for the positive value of the expression \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right) \). This guided us to choose the value of \(k\) that resulted in a positive sum \(k + 1/k\).

It is important to check all possible real roots derived from the substitution to ensure we find the required value of the expression.