Question

Find the zeroes of the quadratic polynomial and verify the relationship between the zeros and the coefficient 
= tsquare-8

Answer 1

Answer:

2√2 and -2√2

Step-by-step explanation:

Given polynomial:

$t^2 - 8\\\\= t^2 - \sqrt{8}^2\\\\= t^2 - (2\sqrt{2})^2\\\\= (t+2\sqrt{2})(t - 2\sqrt{2})$

Thus, the zeroes of the polynomial are 2√2 and -2√2.

Verifying the relationship:-

1. Sum of zeroes = 2√2 - 2√2 = 0; -b/a = -(0)/1 = 0
2. Product of zeroes = 2√2*-2√2 = -8; c/a = -8

Hence, verified.

Answer 2

$\textsf{We have,} \\ \\ \\ \begin{aligned}&t^2-8\\ \\ \Longrightarrow\ \ &t^2+t\sqrt{8}-t\sqrt{8}-8\\ \\ \Longrightarrow\ \ &t(t+\sqrt{8})-\sqrt{8}(t+\sqrft{8})\\ \\ \Longrightarrow\ \ &(t+\sqrt{8})(t-\sqrt{8})\end{aligned}$

$\textsf{So the value of \ t^2-8 \ is \ 0 \ when \ t+\sqrt{8}=0 \ or \ t-\sqrt{8}=0 .} \\ \\ \\ \therefore\ t=\sqrt{8}\ \ \ ; \ \ \ t=-\sqrt{8}\\ \\ \\ \alpha=\sqrt{8}\ \ \ ; \ \ \ \beta=-\sqrt{8}$

$p(t)=t^2-8=0 \\ \\ \\ a=1\ \ \ ; \ \ \ b=0\ \ \ ; \ \ \ c=-8$

$\displaystyle \textsf{Sum of the zeroes:}\\ \\ \\ \alpha+\beta=\sqrt{8}+(-\sqrt{8})=\sqrt{8}-\sqrt{8}=0 \\ \\ \\ -\frac{b}{a}=-\frac{0}{1}=0 \\ \\ \\ \therefore\ \alpha+\beta=-\frac{b}{a}$

$\displaystyle \textsf{Product of the zeroes:}\\ \\ \\ \alpha\beta=\sqrt{8} \times (-\sqrt{8}) = -8 \\ \\ \\ \frac{c}{a}=\frac{-8}{1}=-8\\ \\ \\ \therefore\ \alpha\beta=\frac{c}{a}$

$\huge \textsc{\underline{\underline{Hence Verified!!!}}}$