Understanding the Quadratic Equation Problem

The problem asks us to find one of the possible values for the expression \(x - \frac{1}{{2x}}\) given the quadratic equation \(2x^2 + 5x + 1 = 0\).

We are given the equation: \(2x^2 + 5x + 1 = 0\). We need to evaluate the expression \(x - \frac{1}{{2x}}\).

Manipulating the Given Equation

The equation \(2x^2 + 5x + 1 = 0\) involves terms with \(x^2\), \(x\), and a constant. The expression we need to evaluate, \(x - \frac{1}{{2x}}\), involves \(x\) and \(\frac{1}{x}\) (or \(\frac{1}{{2x}}\)). This suggests we can try to manipulate the given equation to create terms like \(x\) and \(\frac{1}{x}\).

Since \(x=0\) would result in \(2(0)^2 + 5(0) + 1 = 1 \neq 0\), we know that \(x\) is not zero. Therefore, we can safely divide the entire equation by \(2x\):

\(\frac{{2x^2}}{{2x}} + \frac{{5x}}{{2x}} + \frac{1}{{2x}} = \frac{0}{{2x}}\)

This simplifies to:

\(x + \frac{5}{2} + \frac{1}{{2x}} = 0\)

Now, let's rearrange this equation to isolate terms involving \(x\) and \(\frac{1}{{2x}}\):

\(x + \frac{1}{{2x}} = -\frac{5}{2}\)

This gives us the value of the sum \(x + \frac{1}{{2x}}\).

Relating \(x - \frac{1}{{2x}}\) to \(x + \frac{1}{{2x}}\)

We are interested in the expression \(x - \frac{1}{{2x}}\). Let's consider the square of this expression:

\(\left(x - \frac{1}{{2x}}\right)^2 = x^2 - 2(x)\left(\frac{1}{{2x}}\right) + \left(\frac{1}{{2x}}\right)^2\)

\(\left(x - \frac{1}{{2x}}\right)^2 = x^2 - 1 + \frac{1}{{4x^2}}\)

Now, let's consider the square of the sum we found, \(x + \frac{1}{{2x}}\):

\(\left(x + \frac{1}{{2x}}\right)^2 = x^2 + 2(x)\left(\frac{1}{{2x}}\right) + \left(\frac{1}{{2x}}\right)^2\)

\(\left(x + \frac{1}{{2x}}\right)^2 = x^2 + 1 + \frac{1}{{4x^2}}\)

We know that \(x + \frac{1}{{2x}} = -\frac{5}{2}\). Substituting this value into the squared sum equation:

\(\left(-\frac{5}{2}\right)^2 = x^2 + 1 + \frac{1}{{4x^2}}\)

\(\frac{25}{4} = x^2 + 1 + \frac{1}{{4x^2}}\)

From this, we can find the value of \(x^2 + \frac{1}{{4x^2}}\):

\(x^2 + \frac{1}{{4x^2}} = \frac{25}{4} - 1\)

\(x^2 + \frac{1}{{4x^2}} = \frac{25 - 4}{4} = \frac{21}{4}\)

Calculating the Value of \(x - \frac{1}{{2x}}\)

Now we can use the expression for \(\left(x - \frac{1}{{2x}}\right)^2\):

\(\left(x - \frac{1}{{2x}}\right)^2 = x^2 + \frac{1}{{4x^2}} - 1\)

Substitute the value we found for \(x^2 + \frac{1}{{4x^2}}\):

\(\left(x - \frac{1}{{2x}}\right)^2 = \frac{21}{4} - 1\)

\(\left(x - \frac{1}{{2x}}\right)^2 = \frac{21 - 4}{4} = \frac{17}{4}\)

To find \(x - \frac{1}{{2x}}\), we take the square root of both sides:

\(x - \frac{1}{{2x}} = \pm \sqrt{\frac{17}{4}}\)

\(x - \frac{1}{{2x}} = \pm \frac{\sqrt{17}}{\sqrt{4}}\)

\(x - \frac{1}{{2x}} = \pm \frac{\sqrt{17}}{2}\)

So, the two possible values for \(x - \frac{1}{{2x}}\) are \(\frac{\sqrt{17}}{2}\) and \(-\frac{\sqrt{17}}{2}\).

Comparing with Options

Let's look at the given options:

1. \(\frac{{\sqrt {13} }}{2}\)
2. \(\frac{{ {13} }}{2}\)
3. \(\frac{{\sqrt {17} }}{2}\)
4. \(\frac{{ {5} }}{2}\)

One of the values we found, \(\frac{\sqrt{17}}{2}\), matches Option 3.

Revision Table

Additional Information on Solving Quadratic Equations and Expressions

Quadratic equations of the form \(ax^2 + bx + c = 0\) can have real or complex roots depending on the discriminant \(\Delta = b^2 - 4ac\). If \(\Delta > 0\), there are two distinct real roots. If \(\Delta = 0\), there is exactly one real root (a repeated root). If \(\Delta < 0\), there are two distinct complex conjugate roots.

For the given equation \(2x^2 + 5x + 1 = 0\), the discriminant is \(\Delta = 5^2 - 4(2)(1) = 25 - 8 = 17\). Since \(\Delta = 17 > 0\), the equation has two distinct real roots, confirming that our calculations for \(x\) were valid if we had chosen to solve for \(x\) directly.

Manipulating equations by dividing by variables is a common technique when the variable is known to be non-zero. This helps transform the equation into forms that are easier to work with, often involving sums or differences of the variable and its reciprocal, like \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\).

The identity \((a-b)^2 = (a+b)^2 - 4ab\) is a useful general relationship between the square of a difference and the square of a sum. In our case, if we let \(a=x\) and \(b=\frac{1}{{2x}}\), then \((x - \frac{1}{{2x}})^2 = (x + \frac{1}{{2x}})^2 - 4(x)(\frac{1}{{2x}}) = (x + \frac{1}{{2x}})^2 - 2\). This matches our calculation: \(\left(x - \frac{1}{{2x}}\right)^2 = x^2 - 1 + \frac{1}{{4x^2}}\) and \(\left(x + \frac{1}{{2x}}\right)^2 = x^2 + 1 + \frac{1}{{4x^2}}\), so \(\left(x - \frac{1}{{2x}}\right)^2 = \left(x + \frac{1}{{2x}}\right)^2 - 2\). We had \(x + \frac{1}{{2x}} = -\frac{5}{2}\), so \(\left(x - \frac{1}{{2x}}\right)^2 = \left(-\frac{5}{2}\right)^2 - 2 = \frac{25}{4} - 2 = \frac{25-8}{4} = \frac{17}{4}\), leading to the same result \(\pm \frac{\sqrt{17}}{2}\). This confirms the algebraic manipulation is consistent.