Question

if ad not = bc then prove that the equation .(a^2 + b^2)x^2 +2(ac+bd)x +(c^2+d^2) =0 has no equal roots. [note ^ means power and next To^ is its power number)

Answer 1

For a quadratic equation ax^2 + bx + c, the roots are equal if and only if the discriminant is 0. The discriminant is defined as b^2 - 4ac for this case.

In your case, the quadratic equation is {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 + b^2d^2 + a^2d^2 + b^2c^2) 
= -4{(ad)^2 + (bc)^2 - 2*ab*cd} = -4(ad+bc)^2 whcih is zero if and only if ad = bc.

But since that isn't true, the discriminant is non-zero and the roots are unequal.

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  

HOPE THIS ANSWER WILL HELP YOU...