Question

If α and β are the zeroes of the quadratic polynomial p(x) = x²– x – 2, find a polynomial whose zeroes are 2α + 1 and 2β + 1.

{AMU BOARD SAMPLE PAPER CLASS 10th 2017-18}

Answer 1

From the given equation we come to know the following things:
1) The sum of the roots I.e. alpha+beta =1
2) The product of the roots I.e. alpha×beta= -2.
Now we need the equation whose roots are $2 \alpha+1 \ and 2\beta+1$
Hence in the new equation, the sum of the roots and product of roots will become respectively,
$Sum\ of\ the\ roots: 2\alpha +2\beta +2 \\ =2 (\alpha+\beta)+2\\ Product\ of\ roots: (2\alpha+1)(2\beta+1)\\ \Product= 4(\alpha \times \beta) +2 (\alpha+\beta)+1$
By substituting the values of alpha+beta and alpha ×beta got from earlier, we can deduce the new sum of the roots is 4 and the new product of roots is -5.
By using the formula for equation of roots
$x^{2} -(Sum) x +( Product)$ Then we get the required equation as
x^2-4x-5=0

Answer 2

Hi Rehan and others.....

Equation is x² - x - 2

so here , ( α + β ) , i.e sum of zeroes = 1.....(1)

αβ , product = - 2.....(2)


Sum of zeroes of polynomial we needed = 2α + 1 + 2β + 1

= 2( α + β ) + 2

From (1).....

= 2(1) + 2

= 4


product of zeroes we needed = ( 2α + 1 )( 2β + 1 )

= 4αβ + 2α + 2β + 1

= 4αβ + 2( α + β ) + 1

From (1) and (2) ....

= 4(-2) + 2(1) + 1

= -8 + 2 + 1

= -5

So needed polynomial is .....

x² - ( α + β )x + αβ

= x² - 4x - 5.........is your Answer



Hope it helps