Question

$x ^2 - 2 \: find \: the \: zero \: of \: polynomial \: between \: zero \: and \: cofficient$
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Answer 1

Answer:

$x {}^{2} - 2 = 0 \\ x = \sqrt{2} \\ zeros \: of \: the \: given \: polynomial \: \\ = \sqrt{2} \: or \: - \sqrt{2} \\ let \: \alpha \: and \: \beta \: are \: the \:zeros \: of \: \\ given \: polynomial \\ sum \: of \: zeros \: = \alpha + \beta \\ \frac{ - b}{a} = \sqrt{2} + ( - \sqrt{2} ) = \sqrt{2} - \sqrt{2} \\ \frac{ - b}{a} = 0 \\ and \: product \: of \: zeros \: = \alpha \beta \\ \frac{c}{a} = \sqrt{2} \times ( - \sqrt{2} ) \\ \\ \frac{c}{a} = - 2$

Answer 2

Answer:

-√2, √2

Step-by-step explanation:

since x²-2 is a polynomial

to find it's zeroes first factorise

x²-2= (x+√2)(x-√2)

=> zoroes are -√2 and √2