Question

Find the zeroes of xsquare-2 and verify the relationship between zeroes and co effients ..plz solve​

Answer 1

This can be solved by firstly finding the zeros of given polynomial and then by putting values of zeros and coefficients in the relation and it will be verified easily as:-)

Given polynomial is:-

${x}^{2} - 2$

So zeros of the polynomial will be

$\pm \sqrt{2}$

Now relationship between zeros and coefficients.

$sum = \frac{ - coeff \: . \: of \: x}{coeff \: of \: {x}^{2}} \\ so \: \: \: \: - \sqrt{2} + \sqrt{2} = \frac{ - 0}{1} \\ 0 = 0 \: \: \: \: \: \: \: \: \boxed{ \underline{verified}}$

And

$product = \frac{constant}{coeff \: of \: {x}^{2}} \: \\ so \: \: \: \: \: \: - \sqrt{2} \times \sqrt{2} = \frac{ - 2}{1} \\ - 2 = - 2 \: \: \: \: \: \: \: \: \boxed{ \underline{verified}}$

Answer 2

$\large{\mathfrak{\red{\underline{\undderline{........Answer........}}}}}$

=> p(x) = x² - 2

∴ For any zero p(x) = 0

So,

=> x = $\sf{\pm \sqrt{2}}$

So, zero of the polynomial is $\sf{\pm \sqrt{2}}$.

Now, Now relationship between zeros and coefficients:

$\sf{\implies Sum\;of\;zeroes=-\dfrac{b}{a}=\alpha +\beta =-\sqrt{2}+\sqrt{2}=0}$

$\sf{\implies Product\;of\;zeroes=\dfrac{c}{a}=\alpha \beta =-\sqrt{2}\times \sqrt{2}=-2}$

Hence, it is verified!!