Question

if sum and product is -2√3 and -9
respectively find the quadratic polynomial also find zerpes of polynomial

Answer 1

Answer:

The zeroes of the polynomial are -3√3 and √3.

Step-by-step explanation:

Given sum of zeroes = -2√3

Given Product of zeroes = -9

We know that a quadratic polynomial is given by:

k(x² - (Sum of zeroes)x + (Product of zeroes)), where k is the constant term.

So, we have the polynomial

k(x² - (-2√3)x + (-9))

= k(x² + 2√3.x - 9)

Considering k to be 1, we have

Quadratic polynomial = x² + 2√3.x - 9

To obtain its zeroes, we need to split its middle term.

Here, we need -9x² on multiplying and sum as +2√3x.

It can be 3√3.x and -√3.x

By splitting the middle term,

we have x² + 3√3.x + -√3.x - 9

= x(x + 3√3) - √3(x + 3√3)

= (x + 3√3)(x - √3)

To find the zeroes, Let (x + 3√3)(x - √3) = 0

Then, (x + 3√3) = 0 or x = -3√3

And, (x - √3) = 0 or x = √3

Thus, the zeroes of the polynomial are -3√3 and √3.

Answer 2

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♦ As provided

→ Sum of roots = -2√3

→ Product of roots = -9

♦ Now as we know that a quadratic equation is in the form of :-

→ ax² + bx + c

♦ Where

→ $\dfrac{-b}{a} = \textbf{ Sum of roots}$

→ $\dfrac{c}{a} = \textbf{Product of roots}$

♦ So as the value of "a" is not provided we will take it as "1"

♦ Then from above

>> Value of b

$\implies \dfrac{-b}{a} = -2\sqrt{3}$

$\implies \dfrac{-b}{1} = -2\sqrt{3}$

$\implies b = -(-2\sqrt{3})$

$\implies b = 2\sqrt{3}$

>> Value of c

$\implies \dfrac{c}{a} = -9$

$\implies \dfrac{c}{1} = -9$

$\implies c = -9$

♦ Now by substituting the values of a , b and c

$= x^2 + 2\sqrt{3}x - 9$

♦ Now finding out the roots of the equation by splitting the middle term .

$= x^2 + 2\sqrt{3}x - 9$

$= x^2 + 3\sqrt{3}x - \sqrt{3}x - 9$

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

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

♦ So roots are $\sqrt{3} \: and \: -3\sqrt{3}$