if the roots of equation {(ax^2+b^2)x^+2}(bc-ad)x+c^2d^2=0 are equal show that ac+bd=0
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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
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