Question

if the equation (m²+n²)x²-2(mp+nq)x+p²+q²=0 has equal roots, then (a)mp=nq (b)mq=np (c)mn=pq (d)mq=root np​

Answer 1

Answer:

m/n = p/q

Step-by-step explanation:

What is the condition that the equation (m^2+n^2)x^2-2(mp+nq)x + p^2+q^2=0 has equal roots

(m² + n²)x²  - 2(mp + nq)x  + (p² + q²) = 0

for ax² + bx + c = 0

To have equal roots

b² = 4ac

here a = m² + n²  , b = - 2(mp + nq)   c = p² + q²

=> ( - 2(mp + nq))² = 4 (m² + n²)(p² + q²)

=> 4 (m²p² + n²q² + 2mpnq) = 4(m²p² + n²q² + m²q² + n²p²)

Cancelling 4 from both sides

=> m²p² + n²q² + 2mpnq = m²p² + n²q² + m²q² + n²p²

=> m²q² + n²p² - 2mpnq = 0

=> (mq - np)² = 0

=> mq = np

=> m/n = p/q

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Answer 2

$\huge\purple{Answer:-}$

m/n = p/q

Step-by-step explanation:

What is the condition that the equation (m^2+n^2)x^2-2(mp+nq)x + p^2+q^2=0 has equal roots

(m² + n²)x² - 2(mp + nq)x + (p² + q²) = 0

for ax² + bx + c = 0

To have equal roots

b² = 4ac

here a = m² + n² , b = - 2(mp + nq) c = p² + q²

=> ( - 2(mp + nq))² = 4 (m² + n²)(p² + q²)

=> 4 (m²p² + n²q² + 2mpnq) = 4(m²p² + n²q² + m²q² + n²p²)

Cancelling 4 from both sides

=> m²p² + n²q² + 2mpnq = m²p² + n²q² + m²q² + n²p²

=> m²q² + n²p² - 2mpnq = 0

=> (mq - np)² = 0

=> mq = np

=> m/n = p/q