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Question: 48


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Answer 1

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Answer 2

Here is your solution :

For eq'n 1,

⇒ x² + px + q = 0

Let c and d are its zeroes.

Here,

Coefficient of x²( a ) = 1

Coefficient of x( b ) = p

Constant term ( c ) = q

Using Quadratic formula,

⇒ x = [ -b ± √( b² - 4ac ) ] / 2a

⇒ x = [ -p ± √( p² - 4q ) ] / ( 2 × 1 )

⇒ x = [ -p ± √( p² - 4q ) ] / 2

•°• x = [ -p + √( p² - 4q ) ] / 2 = c

And [ -p - √( p² - 4q ) ] / 2 = d

Now,

⇒ c : d = [ -p + √( p² - 4q ) ] / 2 : [ -p - √( p² - 4q ) / 2

•°• c : d = [ -p + √( p² - 4q ) ] : [ -p - √( p² - 4q ) ] ------------- ( 3 )

For eq'n 2,

⇒ x² + bx + c = 0

Let m and n be its zeroes.

Here ,

Coefficient of x² ( a ) = 1

Coefficient of x( b ) = b

Constant term ( c ) = c

Using Quadratic formula,

⇒ x = [ -b ± √( b² - 4ac ) ] / 2a

⇒ x = [ -b ± √( b² - 4c ) ] / 2

•°• x = [ -b + √( b² - 4c ) ] / 2 = m

And,

⇒ [ -b - √( b² - 4c ) ] / 2 = n

Now,

⇒ m : n = [ -b + √( b² - 4c ) ] / 2 : [ -b - √( b² - 4c ) ] / 2

•°• m : n = [ -b + √( b² - 4c ) ] : [ -b - √( b² - 4c ) ] -------- ( 4 )

Now , it is given that ratio of zeroes of both quadratic equation is equal.

So,

⇒ c : d = m : n

Substitute the value of ( 3 ) and ( 4 ),

⇒[ -p + √( p² - 4q ) ] : [ -p - √( p² - 4q ) ] = [ -b + √( b² - 4c ) ] : [ -b - √( b² - 4c ) ]



⇒ [ -p + √( p² - 4q ) ] / [ -p - √( p² - 4q ) ] = [ -b + √( b² - 4c ) ] / [ -b - √( b² - 4c ) ]

We know that,

If ,

⇒ ( a + b ) / ( a - b ) = ( c + d ) / ( c - d )

Then,

⇒ a/b = c/d

⇒( -p ) / [ √( p² - 4q ) ] = ( -b ) / [√( b² - 4c)]


⇒ ( -p ) ÷ ( -b ) = √( p² - 4q ) ÷ √( b² - 4c )

Using Algebric identity,

⇒ √a ÷ √b = √( a ÷ b )

⇒ ( p/b ) = √[ ( p² - 4q ) ÷ ( b² - 4c ) ]

⇒ ( p/b )² = ( p² - 4q ) ÷ ( b² - 4c )

⇒ p² / b² = ( p² - 4q ) ÷ ( b² - 4c )

⇒ p²( b² - 4c ) = b²( p² - 4q )

⇒ p²b² - 4p²c = p²b² - 4b²q

⇒ p²b² - p²b² - 4p²c = -4b²q

⇒ -4p²c = -4b²q

⇒ p²c = ( -4b²q ) ÷ ( -4 )

•°• p²c = b²q

The required answer is option ( 1 ).