Question

If the roots of the equation (a2+b2)x+2 (bc - ad)x + c2 +d2 - O are equal, then prove that ac + BD.​

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

•Answer:

QED

•Step-by-step explanation:

we know that for a Given equation

ax² + 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

Answer:

we know that for a Given equation

ax² + 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