Question

find the zero or zeros of the polynomial and verify the relationship between the zeros and coefficients f(x)=√2x^2+5x√3​

Answer 1

Answer:

The zeros are  $-\sqrt2, \frac{-3}{\sqrt2}$

Step-by-step explanation:

$f(x)=\sqrt2x^2+5x+3\sqrt2$

$f(x)=\sqrt2x^2+2x+3x+3\sqrt2$

$f(x)=\sqrt2x(x+\sqrt2)+3(x+\sqrt2)$

$f(x)=(\sqrt2x+3)(x+\sqrt2)$

The zeros are $-\sqrt2,\frac{-3}{\sqrt2}$

sum of the zeros

$=\frac{-b}{a}$

$=\frac{-5}{\sqrt2}$

product of the zeros

$=\frac{c}{a}$

$=\frac{3\sqrt2}{\sqrt2}$

$=3$

Verification:

sum of the zeros

$=-\sqrt2+\frac{-3}{\sqrt2}$

$=\frac{-2-3}{\sqrt2}$

$=\frac{-5}{\sqrt2}$

product of the zeros

$=(-\sqrt2)(\frac{-3}{\sqrt2})$

$=3$

Hence verified

Answer 2

Step-by-step explanation:

$f(x)=\sqrt2x^2+5x+3\sqrt2f(x)$

$f(x)=\sqrt2x^2+2x+3x+3\sqrt2f(x)$

$f(x)=\sqrt2x(x+\sqrt2)+3(x+\sqrt2)f(x)$

$f(x)=(\sqrt2x+3)(x+\sqrt2)f(x)$

→ Sum of the zeros

$=\frac{-b}{a}</p><p></p><p>[tex]=\frac{-5}{\sqrt2}$

→ Product of the zeros

$=\frac{c}{a}$

$=\frac{3\sqrt2}{\sqrt2}$