Question

find the zeros of the polynomials X2 _3 and verify the relationship between zero and cofficient of the quadratic polynomials ​

Answer 1

$\huge{ \underline{ \sf{ \red{Detailed \: AnsweR\::}}}}$ ‍ ‍ ‍ ‍ ‍ ‍

Given Equation : x² - 3 ‍ ‍ ‍ ‍ ‍ ‍= x² + 0x - 3 ‍ ‍ ‍ ‍ ‍ ‍

To Find : The zeroes of the given polynomial.

In the equation x² + 0x - 3,

$\boxed{ \blue{a = 1}}$

$\boxed{ \green{b = 0}}$

$\boxed{ \purple{c = - 3}}$

We know that, the sum of the zeroes, i.e,

$\alpha + \beta = \frac{ - b}{a}$

$\frac{ - 0}{3} = 0$

And the product of zeroes, i.e,

$\alpha \times \beta = \frac{c}{a}$

$\frac{ - 3}{1} = - 3$

Now simplifying the given polynomial,

x² - 3 = 0x = ±√3

$\sf{Verifying\:The\: Relationship\: Between\: Zeroes\:And\: Coefficients}$

Let us take $\alpha$ as 3 And let us take $\beta$ as -3

Now, $\alpha + \beta = \sqrt{3} + ( - \sqrt{3} ) = 0$

$\alpha \times \beta = \sqrt{3} \times - \sqrt{3} = - 3$

Hence Relation Verified!

Answer 2

Step-by-step explanation:

the quadratic equation is x²+0x-3=0

x²-3=0

x² =3

x. =

+ \sqrt{3} . - \sqrt{3} +

3

.−

3

\alpha + \beta = - b \a \: \:α+β=−b\a

\sqrt{3} + - \sqrt{3} = 0 \1

3

+−

3

=0\1

then 0= 0

\alpha \times \beta = c \a α×β=c\a

\sqrt{3} \times - \sqrt{3} = - 3 \1

3

×−

3

=−3\1

-3=-3