Question

If alpha and beta are the zeroes of a polynomial p (x)= 3x^2-5x + 6 find alpha/beta

Answer 1

Hey There!!

Here, we are given a quadratic polynomial.

$p(x)=3x^2-5x+6$

$\alpha$ and $\beta$ are zeros.

$\text{Sum of zeros } = \alpha+\beta = -\frac{(-5)}{(3)} = \frac{5}{3} \\ \\ \\ \text{Product of zeros } =\alpha \beta = \frac{6}{3} =2$



Also, let 

$\frac{\alpha}{\beta} = t$


Now, consider:

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha \beta} \\ \\ \\ \implies t + \frac{1}{t} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ \\ \\ \implies t + \frac{1}{t} = \frac{(\alpha+\beta)^2}{\alpha \beta}-2$


$\implies \frac{t^2+1}{t} = \frac{\left(\frac{5}{3}\right)^2}{2} -2 \\ \\ \\ \implies \frac{t^2+1}{t} = \frac{25}{18}-2 \\ \\ \\ \implies \frac{t^2+1}{t}=\frac{25-36}{18}$


$\implies \frac{t^2+1}{t} = -\frac{11}{18} \\ \\ \\ \implies 18t^2+11t+1=0 \\ \\ \\ \implies 18t^2+9t+2t+1=0 \\ \\ \\ \implies 9t(2t+1)+1(2t+1)=0 \\ \\ \\ \implies (2t+1)(9t+1)=0 \\ \\ \\ \implies 2t+1=0 \qquad OR \,\,\, 9t+1=0$


$\implies t=-\frac{1}{2} \quad OR t=-\frac{1}{9} \\ \\ \\ \\ \implies \boxed{\frac{\alpha}{\beta}=-\frac{1}{2}} \qquad OR \qquad \boxed{\frac{\alpha}{\beta}=-\frac{1}{9}}$

Hope it helps
Purva
Brainly Community

Answer 2

Hope it helps.............