2. If roots of the quadratic equation x2 + 2px + mn=0 are real and equ

1.	2. If roots of the quadratic equation x2 + 2px + mn=0 are real and equal, show that the roots of thequadratic equation x2 - 2m+n)x+(m? + n2 + 2p) = 0 are also equal.
Answer» x² + 2px + mn = 0Roots are real and equal => D = 0 => (2p)² - 4 (1) (mn) = 0 => 4p² - 4mn = 0 => p² - mn = 0 => p² = mn Substituting p² = mn in the second equation, x² - 2(m + n)x + (m² + n² + 2p²) = 0 => x² - 2(m + n)x + (m² + n² + 2mn) = 0=> x² - 2(m + n)x + (m + n)² = 0 D = [-2(m + n)]² - 4(1)(m + n)² = 4(m + n)² - 4(m + n)² = 0 Since D = 0, the roots are real and equal.

2. If roots of the quadratic equation x2 + 2px + mn=0 are real and equal, show that the roots of thequadratic equation x2 - 2m+n)x+(m? + n2 + 2p) = 0 are also equal.

Answer»

x² + 2px + mn = 0Roots are real and equal => D = 0 => (2p)² - 4 (1) (mn) = 0 => 4p² - 4mn = 0 => p² - mn = 0 => p² = mn

Substituting p² = mn in the second equation, x² - 2(m + n)x + (m² + n² + 2p²) = 0

=> x² - 2(m + n)x + (m² + n² + 2mn) = 0=> x² - 2(m + n)x + (m + n)² = 0

D = [-2(m + n)]² - 4(1)(m + n)² = 4(m + n)² - 4(m + n)² = 0

Since D = 0, the roots are real and equal.