Finding k in a Quadratic Equation with Reciprocal Roots

The question asks us to find the value of the constant \(k\) in the quadratic equation \(5x^2 + 26x + k = 0\), given that one root is the reciprocal of the other.

Understanding the Properties of Quadratic Roots

A general quadratic equation is written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a \neq 0\). For this equation, if the roots are denoted by \(\alpha\) and \(\beta\), there are standard formulas relating the roots to the coefficients:

• Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
• Product of roots: \(\alpha \cdot \beta = \frac{c}{a}\)

Applying Properties to the Given Equation

The given quadratic equation is \(5x^2 + 26x + k = 0\). Comparing this with the standard form \(ax^2 + bx + c = 0\), we can identify the coefficients:

• \(a = 5\)
• \(b = 26\)
• \(c = k\)

We are given that one root is the reciprocal of the other. Let the two roots be \(\alpha\) and \(\beta\). The condition implies that \(\beta = \frac{1}{\alpha}\) (assuming \(\alpha \neq 0\)).

Using the Product of Roots Property

Since we have information about the relationship between the roots through multiplication (reciprocal means their product is 1), the product of roots property is the most helpful here. The product of roots is given by \(\alpha \cdot \beta = \frac{c}{a}\).

Substitute the values of \(a\) and \(c\) from our equation and the condition \(\beta = \frac{1}{\alpha}\):

\(\alpha \cdot \left(\frac{1}{\alpha}\right) = \frac{k}{5}\)

The left side of the equation simplifies:

\(1 = \frac{k}{5}\)

Solving for k

Now, we just need to solve this simple equation for \(k\). Multiply both sides by 5:

\(1 \times 5 = k\)

\(5 = k\)

So, the value of \(k\) is 5.

Verification (Optional)

If \(k=5\), the equation is \(5x^2 + 26x + 5 = 0\). Let's check if the roots are reciprocals. We can solve this quadratic equation. Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(x = \frac{-26 \pm \sqrt{26^2 - 4(5)(5)}}{2(5)}\)

\(x = \frac{-26 \pm \sqrt{676 - 100}}{10}\)

\(x = \frac{-26 \pm \sqrt{576}}{10}\)

Since \(\sqrt{576} = 24\), we have:

\(x = \frac{-26 \pm 24}{10}\)

The two roots are:

• \(x_1 = \frac{-26 + 24}{10} = \frac{-2}{10} = -\frac{1}{5}\)
• \(x_2 = \frac{-26 - 24}{10} = \frac{-50}{10} = -5\)

The roots are \(-\frac{1}{5}\) and \(-5\). Indeed, \(-\frac{1}{5}\) is the reciprocal of \(-5\), since \(-\frac{1}{5} \times (-5) = 1\). This confirms our value of \(k=5\) is correct.

Summary of Finding k

To find \(k\) when one root of \(ax^2 + bx + c = 0\) is the reciprocal of the other, we use the product of roots property: \(\alpha \cdot \beta = \frac{c}{a}\). If \(\beta = 1/\alpha\), then \(\alpha \cdot (1/\alpha) = 1\), so \(1 = \frac{c}{a}\). This means \(c = a\). In our equation \(5x^2 + 26x + k = 0\), we have \(a=5\) and \(c=k\). Setting \(c=a\), we get \(k=5\).

Revision Table: Key Quadratic Concepts

Additional Information: More on Quadratic Roots

The roots of a quadratic equation \(ax^2 + bx + c = 0\) can also be found using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

The term inside the square root, \(b^2 - 4ac\), is called the discriminant, denoted by \(\Delta\) or \(D\). The discriminant tells us about the nature of the roots:

• If \(\Delta > 0\): The roots are real and distinct (different).
• If \(\Delta = 0\): The roots are real and equal (the equation has exactly one solution, often called a repeated root).
• If \(\Delta < 0\): The roots are complex conjugates (not real).

In the case of reciprocal roots, \(\alpha \cdot (1/\alpha) = 1\), which implies \(c/a = 1\), or \(c=a\). This condition is independent of the discriminant and only relates the constant term to the leading coefficient.