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

(Q 23 – 30/4/2 – CBSE 2026 Question Paper)

Ans:

[Approach: We will calculate the value of D and if it is 0, it will be proved that the roots of the given quadratic equation are equal.]

Step 1: Let’s compare the given polynomial with standard quadratic polynomial:

a x ² + b x + c

Here, we get, a = (p – q), b = (q – r) and c = (r – p)

Step 2: Let’s calculate value of discriminant D for the given quadratic equation:

∵ D = b ² – 4 ac …. (i)

By substituting these values in equation (i), we get:

∵ D = (q – r) ² – 4 (p – q)(r – p)

= (q ² + r ² – 2 q r) – 4 (p r – p ² – q r + p q)

= q ² + r ² – 2 q r – 4 p r + 4 p ² + 4 q r – 4 p q

= q ² + r ² – 2 q r + 4 q r – 4 p r + 4 p ² – 4 p q

= (q + r) ² – 4 p (r – p + q)

= (q + r) ² – 4 p (q + r – p) ……… (ii)

Step 3: It is given that, q + r = 2 p

by equation (ii), we get: D = (q + r) ² = 4 p (q + r – p)

we substitute the value of (q + r) in this equation, we get:

D = (2 p) ² – 4 p ((2p) – p)

= 4 p ² – 4 p (p)

= 4 p ² – 4 p ² = 0

∴ D = 0

Step 4: Here, we get value of discriminant as 0,

Hence, it is verified that the roots of the given quadratic equation are equal.

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