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: Notice a special property of the coefficients.
For \((p-q)x^2 + (q-r)x + (r-p) = 0\), add up the three coefficients: \[ (p-q) + (q-r) + (r-p) = 0 \]
Step 2: Use what a zero coefficient sum tells us about the roots.
Whenever the coefficients of a quadratic \(ax^2+bx+c=0\) add up to zero, \(x=1\) is always one of its roots, and the other root is \(\frac{c}{a}\). Here that means one root is \(x=1\) and the other root is \(\frac{r-p}{p-q}\).
Step 3: Use the condition for equal roots.
For both roots to be equal, the second root must also equal 1: \[ \frac{r-p}{p-q} = 1 \implies r - p = p - q \implies q + r = 2p \]
Step 4: Match this with the given condition.
This is exactly the condition \(q + r = 2p\) given in the problem, so whenever \(q+r=2p\) holds, both roots of the equation genuinely become equal to 1. Hence, it is verified that the roots of the given quadratic equation are equal when \(q + r = 2p\).