Question

prove that the equation x^2(a^2+b^2)+2x(ac+bd)+(c^2+d^2)=0 has no real roots if ad is not equal to bc

Answer 1

for no real roots discriminant <0
$dis = \sqrt{ {(2(ac + bd)}^{2} - 4( {d}^{2} + {c}^{2} )( {a}^{2} + {b}^{2}) } \\ dis < 0 \\ (2(ac + bd)) {}^{2} < 4( {d}^{2} {a}^{2} + {d}^{2} {b}^{2} + {c}^{2} {a}^{2} + {b}^{2} {c}^{2} ) \\ {a}^{2} {c}^{2} + {b}^{2} {d}^{2} + 2abcd < {a}^{2} {d}^{2} + {d}^{2} {b}^{2} + {c}^{2} {a}^{2} + {b}^{2} {c}^{2} \\ 2abcd < {a}^{2} {d}^{2} + b {}^{2} {c}^{2} \\ see \: when \: ad = bc \\ lhs = rhs \\ that \: means \: at \: ad = bc \: equation \: will \: have \: roots \: ...but \: for \: rest \: it \: wil \: not \: have \: roots$
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Answer 2

SOLUTION :  

Option (b) is correct :  ad = bc  

Given : (a² + b²)x² - 2(ac + bd)x + ( c² + d²) = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = (a² + b²) , b = - 2( ac + bd)  , c = ( c² + d²)

D(discriminant) = b² – 4ac

Given roots are equal so, D = b² - 4ac = 0

{- 2(ac + bd)}² - 4(a² +b²)(c² + d²) = 0

4(ac + bd)² - 4(a² + b²)(c²+ d²) = 0

4(a²c²+ b²d² + 2abcd ) - 4( a²c² + a²d² + b²d² +  b²c² = 0

[(a + b)² = a² + b² + 2ab]

4(a²c² + b²d² + 2abcd  - a²c² -  a²d² - b²d² -  b²c² ) = 0

(a²c² - a²c² + b²d² - b²d² + 2abcd  -  a²d² -  b²c² ) = 0

2abcd  -  a²d² -  b²c² = 0

-(a²d² + b²c² - 2abcd) = 0  

a²d² + b²c² - 2abcd = 0  

(ad)² + (bc)² - 2×ad × bc = 0

(ad - bc)² = 0

[(a - b)² = a² + b² - 2ab]

ad - bc = 0

ad = bc  

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