Question

Find the zeroes of the polynomial
$t ^{2} - 15$
and verify the relationship be-

tween the zeroes and the coefficients of the polynomial.​

Answer 1

Step-by-step explanation:

$\boxed{Let \: the \: zeroes \: be \: \alpha \: and \: \beta }$

${t}^{2} - 15 = 0$

${t}^{2} = 15$

$t = \pm \sqrt{15}$

$Hence \: the \: roots \: are \: \sqrt{15} \: and \: - \sqrt{15}$

Now we know that,

Sum of its zeroes = -b/a

=> 0/1 = 0

Product of its zeroes = c/a

=> -15/1 = -15

${t}^{2} - ( \alpha + \beta ) x + ( \alpha \times \beta )$

${t}^{2} - (0)x + ( - \sqrt{15} \ )$

${t}^{2} - 15 = 0$

Hence proved.