Question

Quadratic polynomial 2x2-3x+1 has zeros as alpha and beta now form a quadratic polynomial whose zeroes are 3alpha and 3beta

Answer 1

$2 {x}^{2} - 3x + 1 = 0 \\ 2 {x}^{2} - 2x - x + 1 = 0 \\ 2x(x - 1) - 1(x - 1) = 0 \\ (x - 1)(2x - 1) = 0 \\ \alpha = 1 \\ \beta = \frac{1}{2} \\ 3 \alpha = 3 \times 1 = 3 \\ 3 \beta = 3 \times \frac{1}{2} = \frac{3}{2} \\ so \: polynomial \\ {x}^{2} - (3 \alpha + 3 \beta )x + (3 \alpha )(3 \beta ) = 0 \\ {x}^{2} - (3 + \frac{3}{2} )x + (3)( \frac{3}{2} ) = 0 \\ {x}^{2} - \frac{9}{2} x + \frac{9}{2} = 0 \\ 2 {x}^{2} - 9x + 9 = 0$

Answer 2

Answer:2x2-9x+9

Step-by-step explanation:

Alpha + beeta=-b/a

=3/2

Alphabeeta=c/a

=1/2

3alpha +3beeta=3(alpha+beeta)

=3×3/2=9/2=-b/a

3alpha×3beeta=9alphabeeta=9×1/2

=9/2=c/a

a=2 b=-9 c=9

Therefore quadratic polynomial= 2xsquare-9x+9