Question

16.If \alpha and ß are zeroes of the quadratic polynomial
${2y}^{2} - y - 2$
find the
quadratic polynomial whose zeroes are 2 \alpha and 2 ß.​

Answer 1

Answer:

x² - x - 4

Step-by-step explanation:

Quadratic polynomial : 6x² - 7x + 2

α, β are the zeroes of the polynomial

Comparing 2y² - y - 2 with ax² + bx + c we get,

• a = 2
• b = - 1
• c = - 2

Sum of zeroes = α + β = - b / a = - ( - 1 ) / 2 = 1 / 2

Product of zeroes = αβ = c / a = - 2 / 2 = - 1

If 2α , 2β are zeroes of the polynomial

Sum of zeroes = 2α + 2β = 2( α + β ) = 2( 1/2 ) = 1

Product of zeroes = 2α × 2β = 4 × αβ = 4 × ( - 1 ) = - 4

Quadractic polynomial :

= k{ x² - ( Sum )x + Product }

[ Where k ≠ 0 ]

= k{ x² - ( 1 )x + ( - 4 ) }

= k( x² - x - 4 )

When k = 1

= 1( x² - x - 4 )

= x² - 4x - 4

Hence the required quadratic polynomial is x² - x - 4.