Understanding the Quadratic Equation and Roots

The problem presents a quadratic equation in the form \(ax^2+bx+c=0\), specifically \(2x^2-5px+2p^2 = 0\). The core task is to determine the value of the coefficient \(p\), given a specific condition about the equation's roots: one root is 4 units larger than the other.

Analyzing Roots Properties Using Vieta's Formulas

For any quadratic equation \(ax^2+bx+c=0\), let its roots be \(\alpha\) and \(\beta\). Vieta's formulas provide a direct link between the coefficients and the roots:

• Sum of the roots: \(\alpha + \beta = -b/a\)
• Product of the roots: \(\alpha \beta = c/a\)

In our specific equation, \(2x^2-5px+2p^2 = 0\), we identify the coefficients:

• \(a=2\)
• \(b=-5p\)
• \(c=2p^2\)

Applying Vieta's formulas to this equation yields:

• Sum of roots: \(\alpha + \beta = -(-5p)/2 = 5p/2\)
• Product of roots: \(\alpha \beta = (2p^2)/2 = p^2\)

Calculating the Value of p Step-by-Step

The problem states that one root exceeds the other by 4. We can express this relationship mathematically. Let \(\alpha\) be the larger root and \(\beta\) be the smaller root:

\(\alpha = \beta + 4\)

This is equivalent to:

\(\alpha - \beta = 4\)

Now we have a system of equations involving the roots \(\alpha\) and \(\beta\) and the unknown \(p\):

1. \(\alpha + \beta = 5p/2\) (from the sum of roots)
2. \(\alpha - \beta = 4\) (from the given condition)

We can solve this system to find expressions for \(\alpha\) and \(\beta\) in terms of \(p\). Add equation (1) and equation (2):

\((\alpha + \beta) + (\alpha - \beta) = 5p/2 + 4\)

\(2\alpha = 5p/2 + 4\)

Divide by 2 to find \(\alpha\):

\(\alpha = \frac{1}{2} \left( \frac{5p}{2} + 4 \right) = \frac{5p}{4} + 2\)

Next, subtract equation (2) from equation (1):

\((\alpha + \beta) - (\alpha - \beta) = 5p/2 - 4\)

\(2\beta = 5p/2 - 4\)

Divide by 2 to find \(\beta\):

\(\beta = \frac{1}{2} \left( \frac{5p}{2} - 4 \right) = \frac{5p}{4} - 2\)

Now, substitute these expressions for \(\alpha\) and \(\beta\) into the equation for the product of roots (\(\alpha \beta = p^2\)):

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

The left side of the equation is in the form of a difference of squares, \((x+y)(x-y) = x^2 - y^2\), where \(x = 5p/4\) and \(y = 2\). Applying this:

\(\left( \frac{5p}{4} \right)^2 - 2^2 = p^2\)

\(\frac{25p^2}{16} - 4 = p^2\)

To solve for \(p\), rearrange the terms to isolate \(p^2\):

\(\frac{25p^2}{16} - p^2 = 4\)

Combine the \(p^2\) terms using a common denominator:

\(\frac{25p^2}{16} - \frac{16p^2}{16} = 4\)

\(\frac{9p^2}{16} = 4\)

Now, solve for \(p^2\):

\(p^2 = 4 \times \frac{16}{9}\)

\(p^2 = \frac{64}{9}\)

Finally, take the square root of both sides to find the value(s) of \(p\):

\(p = \pm \sqrt{\frac{64}{9}}\)

\(p = \pm \frac{8}{3}\)

The equation yields two possible values for \(p\): \(8/3\) and \(-8/3\). One of these values is \(8/3\).