Question

Find a quadratic polynomial whose zeroes are -9 and -1/9.

Answer 1

$9x^{2}+82x+9$

Tips:

We know that, if $a$ and $b$ are the zeroes of a quadratic polynomial, then the polynomial is given by

$P(x)=(x-a)(x-b)$

$\Rightarrow P(x)=x^{2}-(a+b)x+ab$

Step-by-step explanation:

Given, $(-9)$ and $(-\dfrac{1}{9})$ are the zeroes of a quadratic polynomial.

Then the required quadratic polynomial is given by

$P(x)=[x-(-9)][x-(-\dfrac{1}{9})]$

$\Rightarrow P(x)=(x+9)(x+\dfrac{1}{9})$

$\Rightarrow P(x)=x^{2}+(9+\dfrac{1}{9})x+\dfrac{9}{9}$

$\Rightarrow P(x)=x^{2}+\dfrac{82}{9}x+1$

This quadratic polynomial can be written as

$P(x)=\dfrac{1}{9}(9x^{2}+82x+9)$

$\Rightarrow \boxed{9x^{2}+82x+9}$

#SPJ3

Answer 2

Answer:

$9 x^{2} + 82x + 9$

Step-by-step explanation:

Given: Two zeroes $(-9)$      and $(-\frac{1}{9})$ of a quadratic equation

To Find: The quadratic equations with these zeroes

Concept: If $a$ and $b$ are zeroes of a quadratic polynomial then the polynomial can be written as :

$Q(x) = (x - a)(x - b)$

≅ $Q(x) = x^{2} - (a+b)x + ab$

Solution:

Here, zeroes are (-9) and $(-\frac{1}{9})$,

So,  $a = (-9)$ and $b = (-\frac{1}{9} )$

putting the values in above equation,

$Q(x) = [x - (-9)][x - (-\frac{1}{9})]$

⇒ $Q(x) = [x + 9][x + \frac{1}{9}]$

⇒ $Q(x) = x^{2} + (9 + \frac{1}{9})x + (9 * \frac{1}{9})$

⇒ $Q(x) = x^{2} + \frac{82}{9}x + 1$

⇒ $Q(x) = \frac{1}{9}(9 x^{2} + 82x + 9)$

∴ the required polynomial is $9 x^{2} + 82x + 9$

#SPJ2