Question

Verify that roots of the quadratic equation \((p - q)x^2 + (q - r)x + (r - p) = 0\) are equal when \(q + r = 2p\).

Hint:

Notice that the sum of the coefficients of the given quadratic equation is:
\[ a + b + c = (p - q) + (q - r) + (r - p) = 0 \] Whenever the sum of the coefficients of a quadratic equation \(ax^2 + bx + c = 0\) is zero, one of its roots is always \(x = 1\), and the other root is \(\frac{c}{a}\).
For the roots to be equal, we must have:
\[ \frac{c}{a} = 1 \implies c = a \] \[ r - p = p - q \implies q + r = 2p \] Using this root-sum property allows you to solve and verify the condition in just two lines!

Answer 1

Step 1: Understanding the Question:
The topic of this question is Quadratic Equations.
A quadratic equation of the form \(ax^2 + bx + c = 0\) has real and equal roots if and only if its discriminant \(D = b^2 - 4ac\) is exactly equal to zero.
We are given the quadratic equation \((p - q)x^2 + (q - r)x + (r - p) = 0\), where the coefficients themselves are expressions involving the real numbers \(p\), \(q\), and \(r\).
We need to show that the roots of this quadratic equation are equal when the algebraic condition \(q + r = 2p\) is satisfied.

Step 2: Key Formula or Approach:
Let us compare the given quadratic equation with the standard quadratic equation \(ax^2 + bx + c = 0\):
Here, the coefficients are:
\(a = p - q\)
\(b = q - r\)
\(c = r - p\)
The discriminant of this quadratic equation is given by:
\[ D = b^2 - 4ac = (q - r)^2 - 4(p - q)(r - p) \] We will substitute the given condition \(q + r = 2p\) (or \(2p = q + r\)) into the expression for the discriminant and simplify it to show that \(D = 0\).

Step 3: Detailed Explanation:

• Express the given condition in terms of \(p\):
Since \(q + r = 2p\), we can write:
\[ p = \frac{q + r}{2} \]

• Substitute this expression for \(p\) into the coefficient \(a = p - q\):
\[ a = \frac{q + r}{2} - q \] \[ a = \frac{q + r - 2q}{2} \] \[ a = \frac{r - q}{2} \]

• Substitute this expression for \(p\) into the coefficient \(c = r - p\):
\[ c = r - \frac{q + r}{2} \] \[ c = \frac{2r - (q + r)}{2} \] \[ c = \frac{r - q}{2} \]

• Notice that the third coefficient \(b\) can be written as:
\[ b = q - r = -(r - q) \]

• Substitute these simplified expressions for \(a\), \(b\), and \(c\) into the discriminant formula \(D = b^2 - 4ac\):
\[ D = (q - r)^2 - 4\left(\frac{r - q}{2}\right)\left(\frac{r - q}{2}\right) \] \[ D = (r - q)^2 - 4\left(\frac{(r - q)^2}{4}\right) \]

• Simplify the terms inside the expression:
The factors of 4 in the numerator and denominator cancel out:
\[ D = (r - q)^2 - (r - q)^2 \] \[ D = 0 \] Since the discriminant \(D\) simplifies to exactly zero under the given condition, the roots of the quadratic equation must be real and equal.


Step 4: Final Answer:
Hence, it is verified that the roots of the given quadratic equation are equal when \(q + r = 2p\).