Question

if the zeroes of ttthe polynomial x^3-9x^2+3x+1are alfa-bita,alfa,alfa+bita,then the value of alfa is ​

Answer 1

Answer:

$\boxed{\bf{=> \alpha = 3}}$

Step-by-step explanation:

$\text{Given polynomial,} \\ {\bf => x^3-9x^2+3x+1 }\\\\ \text{It is of the form }ax^3+bx^2+cx+ d\\ a =1,b=-9,c=3,d=1 \\\\ \alpha -\beta,\alpha,\alpha+\beta \ \text{are the zeroes of the given polynomial.} \\\\ => \boxed{\bf{sum \ of \ zeroes = \frac{-b}{a}}} \\\\ \alpha -\beta+\alpha+\alpha+\beta = \frac{-(-9)}{1} \\\\ 3\alpha = 9 \\\\ \alpha=\frac{9}{3} \\\\ \boxed{\bf \alpha = 3 }$