Math, asked by CuteSiddhi, 1 year ago

pls answer it....urgent....Q21(ii)part

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Answered by siddhartharao77
2
(ii)

Given Equation is (a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0

Given that the Equation has no real roots.

D = b^2 - 4ac < 0

= > 2(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2)

= > 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2)

= > 4a^2c^2 + 4b^2d^2 + 8abcd - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2

= > 8abcd - 4a^2d^2 - 4b^2c^2

= > -4(ad - bc)^2


The value of D is -ve. The roots cannot be -ve. Therefore the equation has no real roots. 


Hope this helps!
Answered by Anonymous
1
Here is your solution :

Given,

=> ( a² + b² )x² + 2( ac + bd ) + ( c² + d² ) = 0

Here,

Coefficient of x²( a ) = ( a² + b² )

Coefficient of x( b ) = 2( ac + bd )

Constant term ( c ) = ( c² + d² )

Now, for any quadratic equation to have no real zeroes its Discriminat ( D = b² - 4ac ) should be less than ( < ) 0.

=> D < 0

=> b² - 4ac < 0

Substitute the value of a,b and c.

=> ( 2ac + 2bd )² - 4( a² + b² ) ( c² + d² ) < 0

Using identity:

[ ( a + b )² = a² + b² + 2ab ]

=> ( 2ac )² + ( 2bd )² + 2( 2ac )( 2bd ) - 4 ( a²c² + a²d² + b²c² + b²d² ) < 0

=> 4a²c² + 4b²d² + 8abcd - 4a²c² - 4a²d² - 4b²c² - 4b²d² < 0

=> 8abcd - 4a²d² - 4b²c² < 0

Taking 4 as common,

=> 4 ( 2abcd - a²d² - b²c² ) < 0

=> 2abcd - a²d² - b²c² < 0 / 4

=> 2abcd - a²d² - b²c² < 0

=> -( a²d² + b²c² - 2abcd ) < 0

=> ( a²d² + b²c² - 2abcd ) < 0

=> ( ad )² + ( bc )² - 2( ad ) ( bc ) < 0

Using identity :

[ ( a² + b² - 2ab ) = ( a - b )² ]

=> ( ad - bc )² < 0

=> ( ad - bc ) < 0

•°• ad < bc

Here, ad ≠ bc i.e. ad < bc and it is having no real zeroes.
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