Question

please answer first 4 questions

I will mark you as brainiest if you gave quality answer

these are class 10th questions.

Answer 1

Hey Mate !

Here is your solution :

1.

Here,we have to find the value of k such that the lines of both equations are coincident.

It means that the equation should be have infinitely many solutions.

Here we go ,

=> ( k + 1 )x + 3ky + 15 = 0 -------- ( 1 )

Here,

=> a1 ( Coefficient of x ) = ( k + 1 )

=> b1 ( Coefficient of y ) = 3k

=> c1 ( Constant term ) = 15

" And "

=> 5x + ky + 5 = 0 ----------- ( 2 )

Here,

=> a2 ( Coefficient of x ) = 5

=> b2 ( Coefficient of y ) = k

=> c2 ( Constant term ) = 5

For Pair of Linear equation in two variables to have infinitely many solutions,

=> a1/a2 = b1/b2 = c1/c2


=> ( k + 1 ) / 5 = 3k / k = 15 / 5


=> ( k + 1 ) / 5 = 3 = 3

=> ( k + 1 ) / 5 = 3

=> ( k + 1 ) = 3 × 5

=> ( k + 1 ) = 15

=> k = 15 - 1

•°• k = 14

2.

=> ( 4/x ) + 5y = 7 --------- ( 1 )

=> ( 3/x ) + 4y = 5 ---------- ( 2 )

Multiplying in ( 1 ) by 3 and in ( 2 ) by 4 ,

=> 3 [ ( 4/x ) + 5y ] = 7 × 4

=> ( 12/x ) + 15y = 28 --------- ( 3 )

" And "

=> 4 [ ( 3/x ) + 4y ] = 5 × 4

=> ( 12/x ) + 16y = 20 ---------- ( 4 )

Subtract ( 3 ) from ( 4 ),

=> ( 12/x ) + 16y - [ ( 12/x ) + 15y ] = 20 - 28


=> ( 12/x ) + 16y - ( 12/x ) - 15y = -8

=> 16x - 15y = -8

•°• x = -8

Substitute the value of x in ( 1 ),

=> ( 4/x ) + 5y = 7

=> [ 4/( -8 ) ] + 5y = 7

=> ( 1/-2 ) + 5y = 7

=> ( -1/2 ) + 5y = 7

=> 5y = 7 + ( 1/2 )

=> 5y = ( 14 + 1 ) / 2

=> 5y = 15/2

=> y = 15/ ( 2 × 5 )

•°• y = 3/2


Hence,

=> x = -8

=> y = 3/2

3.

Given,

Zeroes = ( 2 + √3 ) and ( 2 - √3 )

Now,

=> Sum of zeroes = 2 + √3 + 2 - √3

•°• Sum of zeroes = 4

" And "

=> Product of zeroes = ( 2 + √3 ) ( 2 - √3 )

Using identity :

[ ( a + b ) ( a - b ) = ( a² - b² )

=> Product of zeroes = ( 2 )² - ( √3 )²

=> Product of zeroes = 4 - 3

•°• Product of zeroes = 1

The formula for a quadratic polynomial is

=> x² - ( Sum of zeroes )x + Product of zeroes = 0

=> x² - ( 4 )x + 1 = 0

=> x² - 4x + 1 = 0

The required quadric polynomial is :

=> x² - 4x + 1 = 0

4.

=> px - qy + r = 0 ------- ( 1 )

Here,

=> a1 ( Coefficient of x ) = p

=> b1 ( Coefficient of y ) = -q

=> c1 ( Constant term ) = r

" And "

=> ax + by + c = 0 -------- ( 2 )

Here,

=> a2 ( Coefficient of x ) = a

=> b2 ( Coefficient of y ) = b

=> c2 ( Constant term ) = c

To have unique solution for a pair of linear equation in two variables,

=> a1/a2 ≠ b1/b2

•°• p / a ≠ -q / b

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Hope it helps !! ^_^