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)
Answers & Comments
BeUnknown
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.
Answers & Comments
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.
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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