Question

If a, ß are zeroes of x2 – 11x+28 then
a B+ a B2
(​

Answer 1

Answer:

65

Step-by-step explanation:

$\alpha + \beta = 11 \\ \alpha \beta = 28 \\ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ = {11}^{2} - 2 \times 28 \\ = 121 - 56 \\ = 65$

Aliter method

${x}^{2} - 11x + 28 \\ {x}^{2} - 7x - 4x + 28 \\ x(x - 7) - 4(x - 7) \\ (x - 7)(x - 4) \\ zeroes \: are \: 7 \: and \: 4 \\ so \: let \: \alpha = 7 \\ \beta = 4 \\ { \alpha }^{2} + { \beta }^{2} = {7}^{2} + {4}^{2} = 49 + 16 = 65$

Answer 2

Step-by-step explanation:

since ab = c/a then

28=14 and (x-7)(x-4)then

x=4,7then

ans is 196