Question

If the roots of the equation (a²+ b²)x² − 2 (ac + bd)x + (c² + d²) = 0 are equal, prove that $\frac{a}{b}=\frac{c}{d}$.

Answer 1

SOLUTION :  

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]

Square root both sides,  

ad - bc = 0

ad = bc  

a/b = c/d

Hence, proved a/b = c/d

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Answer 2

 It is given:

(a2 + b2)x2 - 2(ac+bd)x + (c2 + d2) = 0

To prove:

a / b = c / d

PROOF:

we know that

D = b2 - 4ac = 0 (for equal roots)

b2 = 4ac

{-2(ac + bd)}2 = 4{(a2 + b2) (c2 + d2)}

4(a2c2 + b2d2 +2acbd) = 4(a2c2 + a2d2 + b2c2 + b2d2)

2acbd = a2d2 + b2c2

a2d2 + b2c2 - 2abcd = 0

(ad - bc)2 = 0

Taking square root on both sides

ad - bc = 0

ad = bc

a / b = c / d

hence proved.

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