Question

There are no results for 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.

Answer 1

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.

Answer 2

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• 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

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