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:

An alternative way to solve this is to substitute \(p = \frac{q + r}{2}\) directly into the discriminant formula \(D = B^2 - 4AC\) and show that it reduces to zero.
However, utilizing the root properties of equations with coefficient sums equal to zero is much faster and reduces algebra steps!

Answer 1

Step 1: Simplify $A=p-q$ and $C=r-p$ using the given condition.
Here $A=p-q$, $B=q-r$, $C=r-p$, and $q+r=2p \implies p=\frac{q+r}{2}$.
Step 2: Substitute to find A and C.
$A=p-q=\frac{q+r}{2}-q=\frac{r-q}{2}$, and $C=r-p=r-\frac{q+r}{2}=\frac{r-q}{2}$, so $A=C=\frac{r-q}{2}$.
Step 3: Compute the discriminant directly.
$D=B^2-4AC=(q-r)^2-4\left(\frac{r-q}{2}\right)^2=(r-q)^2-(r-q)^2=0$, using $(q-r)^2=(r-q)^2$.
Step 4: Conclude.
Since the discriminant is exactly zero, the roots of the given equation are equal whenever $q+r=2p$. Hence verified.