Question

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Answer 1

Heya!!

Here is your solution :

Question no.8

Given,

α and ß are zeroes of quadratic polynomial ( x² - 6x + k ).

Here,

⇒Coefficient of x² ( a ) = 1

⇒Coefficient of x ( b ) = -6

⇒Constant term ( c ) = k

And,

⇒3α + 2ß = 20

⇒3α = 20 - 2ß

•°• α = ( 20 - 2ß )/3 ------------ ( 1 )

Now,

We know the relationship between zeroes and coefficient of a quadratic polynomial.

i.e.

⇒Sum of zeroes = ( -b / a )

⇒α + ß = [ - ( -6 ) / 1 ]

⇒α + ß = 6

•°• α = 6 - ß -------------- ( 2 )

From ( 1 ) and ( 2 ), we get ,

⇒ 6 - ß = ( 20 - 2ß ) / 3

⇒3( 6 - ß ) = ( 20 - 2ß )

⇒18 - 3ß = 20 - 2ß

⇒18 - 20 = 3ß - 2ß

•°• ß = -2

Substitute the value of ß in ( 2 ),

⇒α = 6 - ß

⇒α = 6 - ( -2 )

⇒ α = 6 + 2

•°• α = 8

Now,

⇒Product of zeroes = c/a

⇒αß = k/1

⇒αß = k

Plug the valuee of α and ß ,

⇒( -2 ) 8 = k

•°• k = -16

( a ) -16 ✓✓


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Question no.9

Given,

One of the zeroes of quadratic polynomial ( 2x² - 3x + k ) is reciprocal to other.

Here,

⇒Coefficient of x² ( a ) = 2

⇒Coefficient of x ( b ) = -3

⇒Constant term ( c ) = k

Let one of the zeroes of this quadratic polynomial is ß, so other zero will be ( 1/ß ).

Now,

We know the relationship between zeroes and coefficient of a quadratic polynomial.

⇒Product of zeroes = c/a

⇒ß ( 1/ß ) = k/2

⇒1 = k/2

⇒k = 1 × 2

•°• k = 2

( a ) 2 ✓✓

Answer 2

$8. \: {x}^{2} - 6x + k = 0 \\ \\ Sum \: \: of \: \: zeroes = \frac{ - ( - 6)}{1} = 6 \\ \\ \alpha + \beta = 6 \\ 3 \alpha + 2 \beta = 20 \\ \\ Solving \: \: both \: \: equations \: \: , \\ We \: \: get \: \: - \\ \\ \alpha = 8 \: \: \: \: \: \: \: \: \beta = - 2 \\ \\ Now \: \: , \: \: Product \: \: of \: \: roots \: = \frac{k}{1} \\ \\ || \: \: \: k = - 16 \: \: \: ||$



$9. \: 2 {x}^{2} - 3x + k = 0 \\ \\ Since \: the \: roots \: are \: reciprocal \: \\ of \: each \: other \: \\ say \: \: \alpha \: \: and \: \: \dfrac{1}{ \alpha } \: \\ \\ Product \: of \: roots \: = \: 1 = \dfrac{k}{2} \\ \\ Therefore \: , \: \: || \: \: k = 2 \: \: ||$