Question

find the zeros of polynomial 2x square-9.

Answer 1

Answer:

The answer is $x=\frac{-3}{\sqrt{2} } or x=\frac{3}{\sqrt{2} }$

step-by-step explanation:

The zeros of a polynomials, p(x), all seem to be x-values that make it equal to zero. They are intriguing to us for a number of reasons, one of which is that they provide information on the graph's x-intercepts for the polynomials. They are directly related to the parameters of the quadratic, as we will see see.

A number "a" such that p(a) = 0 is the value zero of the polynomial p(x). If p(x) is partitioned by the sequential polynomial x - a, then the result is p, where p(x) is a polynomial of degree higher than or equal to 1 and an is any true figure (a).

Let the polynomial equation's solutions be, α,β.

Let the roots of the quadratic equation be $\alpha ,\beta =2x^{2} +0x-9=0$

$(\sqrt{2x} )^{2} -(3)^{2} =0$

$(\sqrt{2x}+3 )(\sqrt{2x}-3 )=0$

$\sqrt{2x} +3=0$       $or$            $\sqrt{2x} -3=0$

$x=\frac{-3}{\sqrt{2} }$                $or$           $x=\frac{3}{\sqrt{2} }$

Thus, zeros of the polynomial $2x^{2} -9$ is $x=\frac{-3}{\sqrt{2} } orx=\frac{3}{\sqrt{2} }$

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