If the roots of the equation, (a 2 +b 2 )x 2 -2(ac+bd)x+(c 2 +d 2 ) = 0 are equal the
prove that ad = bc.
Answers
Note:
• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .
• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.
• The discriminant of the the quadratic equation
Ax² + Bx + C = 0 , is given as ; D = B² - 4A•C
• If D > 0 then its roots are real and distinct.
• If D < 0 then its roots are imaginary.
• If D = 0 then its roots are real and equal.
Solution:
Here,
The given quadratic equation is :
(a² + b²)x² - 2(ac + bd)x + (c² + d²) = 0
Clearly, here we have ;
A = a² + b²
B = -2(ac + bd)
C = c² + d²
Now,
The discriminant will be ;
=> D = B² - 4A•C
=> D = [-2(ac + bd)]² - 4(a² + b²)•(c² + d²)
=> D = 4•(a²c² + b²d² + 2abcd)
– 4•(a²c² + a²d² + b²c² + b²d²)
=> D = 4•[ (a²c² + b²d² + 2abcd )
– (a²c² + a²d² + b²c² + b²d²) ]
=> D = 4•[ a²c² + b²d² + 2abcd
– a²c² – a²d² – b²c² – b²d² ]
=> D = 4•[ 2abcd – a²d² – b²c² ]
=> D = -4•[ a²d² + b²c² - 2abcd ]
=> D = -4•[ (ad)² - 2•ad•bc + (bc)² ]
=> D = -4•[ ad - bc ]²
It is given that,
The roots of the given quadratic equation are equal, thus the discriminant D must be equal to zero.
Thus,
=> D = 0
=> -4•[ ad - bc ] = 0
=> ad - bc = 0
=> ad = bc
Hence proved.
- If roots are equal , then ∆ = 0
Given equation,
- (a²+b²)x² - 2(ac+bd)x + (c²+d²) = 0
Calculating the value of ∆
- ∆ = (2(ac+bd))² - 4(a²+b²)(c²+d²) = 0
- 4(ac)²+4(bd)²+8acbd -4(ac)² -4(ad)² -4(bc)² -4(bd)² = 0
- 8abcd - 4(ad)² -4(bc)² = 0
- -4(ad-bc)² = 0
- ad - bc = 0
- ad = bc