Question

if the roots of the equation (a2+B2)X2+2(BC-AD)X+c2+d2=0 are real and equal show that AC+BD=0
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Answer 1

Answer:

QED

Step-by-step explanation:

we know that for a Given equation

ax^2 + bx + c = 0

roots are equal & Real when

d = b² - 4ac = 0

for given quation

b = 2(BC-AD)

a = A² + B²

c = C² + D²

putting these values we get

(2(BC -AD))² = 4(A² + B²)(C² + D²)

=> 4(BC -AD)² = 4(A² + B²)(C²) + (A² + B²)(D²)

Cancelling 4 from both sides

=> (BC -AD)² =(A² + B²)(C²) + (A² + B²)(D²)

Expanding square

=> (BC)² + (AD)²  -2BCAD = A²C² +B²C²  + A²D² + B²D²

=> B²C² + A²D² - 2ACBD = A²C² +B²C²  + A²D² + B²D²

cancelling B²C² + A²D²  from both sides

=>  - 2ACBD = A²C²  + B²D²

=> 0 = A²C²  + B²D² + 2ACBD

=> A²C²  + B²D² + 2ACBD = 0

=> (AC + BD)² = 0

=> AC +BD = 0

QED

Answer 2

Step-by-step explanation:

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