Question

if the equation (m^2+n^2)x^2-2(mp+nq)x+p^2+q^2=0 has equal roots, then
a) mp=nq
b) mq=np
c) mn=pq
d) mq=√np​

Answer 1

Answer:

(m²+ n²)x2 - 2 x (mp + nq) +(p²+q²)=0

for equal roots condition is D=0

D = b² - 4ac

b² = 4 ac

[2(mp + nq)]² = 4 (m²+ n2)(p² + q²)

4 [ (mp)² + (nq)² + 2mp nq] = 4 [(mp)² + (mq)² + (np)² + (nq)²]

(mp)²+(nq)² + 2 mp nq = (mp)² + (mq)² + (np)² + (nq)²

2 mp bq = (mq)² + (np)²

0= (mq)² + (np)² - 2 mp nq

using (a-b)² = a² + b² - 2ab

0 =( mq - np)²

mq - np = 0

mq = np

therefore, correct answer is (b) mq = np

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Answer 2

Answer:

(m²+ n²)x2 - 2 x (mp + nq) +(p²+q²)=0

for equal roots condition is D=0

D = b² - 4ac

b² = 4 ac

[2(mp + nq)]² = 4 (m²+ n2)(p² + q²)

4 [ (mp)² + (nq)² + 2mp nq] = 4 [(mp)² + (mq)² + (np)² + (nq)²]

(mp)²+(nq)² + 2 mp nq = (mp)² + (mq)² + (np)² + (nq)²

2 mp bq = (mq)² + (np)²

0= (mq)² + (np)² - 2 mp nq

using (a-b)² = a² + b² - 2ab

0 =( mq - np)²

mq - np = 0

mq = np

therefore, correct answer is (b) mq = np

Hope it may help you........

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