Question

(a^2+b^2)x^2+2(bc-ad)x+c^2+d^2=0 than p t ac+bd=0

Answer 1

$\huge{\underline{\sf{Solution:-}}}$

$\bigstar\underline{\sf{Given:-}}$

• ax² + bx + c = 0

$\huge{\underline{\sf{Explaination:-}}}$

d = b² - 4ac = 0

b = 2(BC - AD)

a = A² + B²

c = C² + D²

$\bigstar\underline{\sf{Putting\:these\:values\:we\:get:-}}$

(2(BC - AD))² = 4(A² + B²)(C² + D²)

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

$\bigstar\underline{\sf{Cancelling\:4\:from\;both\;sides:-}}$

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

$\bigstar\underline{\sf{Expanding\:square:-}}$

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

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

$\bigstar\underline{\sf{Cancelling\;B^{2}C^{2} + A^{2}D^{2}\:from\:both\:sides:-}}$

=> - 2ACBD = A²C² + B²D²

=> 0 = A²C² + B²D² + 2ACBD

=> A²C² + B²D² + 2ACBD = 0

=> (AC + BD)² = 0

=> AC + BD = 0

______________________________________________________________________________

Answer 2

$\huge{\underline{\bf{Answer:-}}}$

★ Given:-

• ax² + bx + c = 0

Roots are equal & Real when

• d = b² - 4ac = 0

$\huge{\underline{\bf{Explainationtion:-}}}$

Given equation:-

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