Question

Please tell 11th answer. CLASS 10​

Answer 1

Answer:

${x}^{2} - 6x - 27 \\ = {x}^{2} + 3x - 9x - 27 \\ = x(x + 3) - 9(x + 3) \\ = (x - 9)(x + 3) \\ x = 9 \: or \: x = - 3$

If the two roots were alpha and beta, their relations with the coefficients would be as follows:

$\alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 6)}{1} = \frac{6}{1} = 6 \\ \alpha \times \beta = \frac{c}{a} = \frac{ - 27}{1} = - 27$

Hope this helps.

Thanks.

Answer 2

Step-by-step explanation:

x²-6x-27=0

=x²-9x+3x-27=0

=x(x-9)+3(x-9)=0

=(x-9)(x+3)=0

=>x=9OR x=-3

sum of zeros=9+(-3)=-b/a

=-6

product of zeros=9×(-3)=c/a

=-27