Question

CLASS X MATHS...


1.If a and B are two zeroes of the polynomial x²-2x-15 then form a quadratic polynomial
whose zeroes 2a and 2B.


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Answer 1

Answer:

Given : If α and β are zeroes of the polynomial x²-2x-15

Need To Find : A quadratic polynomial whose zeroes 2α and 2β.

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❍ First Find out all Cofficients of Given Polynomial x²-2x-15 :

• Cofficient of x² = 1 .
• Cofficient of x = -2 .
• Constant Term = -15 .

⠀⠀⠀⠀⠀ As , We know that α and β are two zeroes of Polynomial x²-2x-15 .

As, We know that :

⠀⠀⠀⠀⠀$\implies \sf{\alpha + \beta = \dfrac{- Cofficient \:of\:x \:}{Cofficient \:of\:x^{2} }}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Cofficients\::}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies \tt{\alpha + \beta = \dfrac{-(-2) \:}{1 }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies \tt{\alpha + \beta = \dfrac{2 \:}{1 }}$

Or ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies\boxed {\sf{\alpha + \beta = 2 \:}}\:\bf{\star}$

As, We know that :

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\implies \sf{\alpha \times \beta = \dfrac{Constant\:Term \:}{Cofficient \:of\:x^{2} }}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Cofficients\::}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies \tt{\alpha \times \beta = \dfrac{-15 \:}{1 }}$

Or ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies\boxed {\sf{\alpha \times \beta = -15 \:}}\:\bf{\star}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

As, We know that ,

• Quadratic Polynomial = x² - (sum of zeroes ) x + Product of zeroes

⠀⠀⠀⠀⠀

Given that :

• The zeroes of new formed polynomial will be 2α and 2β .

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$:\implies {\tt{ \bigg( x^{2} - ( 2\alpha + 2\beta ) x+ ( 2\alpha \times 2 \beta ) \bigg) }}$

⠀⠀⠀⠀⠀⠀$:\implies {\tt{ \bigg( x^{2} -2 ( \alpha + \beta ) x +( 2\alpha \times 2 \beta ) \bigg) }}$

⠀⠀⠀⠀⠀⠀$:\implies {\tt{ \bigg( x^{2} -2 ( \alpha + \beta ) x + 4\alpha\beta \bigg) }}$

As , We have found ,

⠀$\implies \tt{\alpha \times \beta = -15 \:}$

⠀$\implies \tt{\alpha + \beta = 2 \:}$

⠀⠀⠀⠀⠀⠀$:\implies {\tt{ \bigg( x^{2} -2 ( 2 ) x + 4(-15) \bigg) }}$

⠀⠀⠀⠀⠀⠀$:\implies {\tt{ \bigg( x^{2} -4x -60 \bigg) }}$

Or ,

⠀⠀⠀⠀⠀⠀$:\boxed{\sf{ New\:Formed \:Quadratic \:Polynomial = x^{2} -4x -60 }}$

⠀⠀⠀⠀⠀

Therefore,

⠀⠀⠀⠀⠀⠀$\therefore \underline{\sf{ New\:Formed \:Quadratic \:Polynomial = x^{2} -4x -60 }}$

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