Question

find the zeroes of the polynomial x2 - 3 and verify the relationship between the zeroes and the coefficient​

Answer 1

Step-by-step explanation:

x^2-3=0

x^2=3

then,

$x = ( - \sqrt{3} ) \: \: or( + \sqrt{3)}$

coefficient of x^2=1

Answer 2

Step-by-step explanation:

${x}^{2} - 3$

WHENEVER THE SECOND TERM IS MISSING WE HAVE TO USE THE FORMULA

${a}^{2} - {b}^{2}$

$= (a + b) \: ( a- b)$

SOLVING IT...

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

WE HAVE GOT 2 ZEROS

VERIFYING THE RELATIONSHIP:-

$sum \: of \: the \: zeroes \: \\ \alpha + \beta = \frac{ - b}{a} \\ \sqrt{3} + ( - \sqrt{3} ) = \frac{ - b}{a} \\ 0 = 0 \\ product \: of \: the \: zeroes \\ \alpha \beta = \frac{c}{a} \\ \sqrt{3} \times ( - \sqrt{3}) = \frac{c}{a} \\ (- 3) = (- 3)$

HERE YOU GO...