Math, asked by modleraditya8988, 9 months ago

If m and n are real and
different satisfying the relation
m2 - 4m +4
and
n² = 40-4
, then
(m-* - 4mm)
is equal to:​

Answers

Answered by 165
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

Answered by akshara0108
0

Answer:

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Step-by-step explanation:

m/n = p/q

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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