Question

If ad is not equal to bc, then prove that the equation (a^2+b^2)x^2+ 2( ac+bd)x +(c^2+d^2)=0 has real roots

Answer 1

Its an equation in x
So,
For the roots to be real b^2-4ac > 0
Now,
[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

So it is more than
So roots are real

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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