Question

4. If the sum of the zeroes is -p and product of the zeroes is -1/p, then the quadratic polynomial is​

Answer 1

Answer:

x²+p-(1/p)

Step-by-step explanation:

The general form of a quadratic equation is: x²-(α+β)x+(αβ)

Given:

α+β=-p

αβ=-1/p

∴The quadratic equation is: x²-(-p)+(-1/p)

=x²+p-(1/p) is the required quadratic equation.

Answer 2

Given:

Sum of zeroes = -p

Product of zeroes = $\frac{-1}{p}$

To find:

The quadratic polynomial.

Solution:

Let $\alpha$ be one of the zeroes and $\beta$ be the other zero. Then sum of zeroes, $\alpha +\beta =-p$

Product of zeroes, $\alpha \beta =\frac{-1}{p}$

Consider a quadratic equation, $ax^{2} +bx+c=0$

Then, the sum and product of zeroes in the quadratic equation will be:

$x^{2} -(\alpha +\beta)x+(\alpha \beta )=0$

Here, $\alpha +\beta =-p$ and $\alpha \beta =\frac{-1}{p}$

Substituting in the quadratic equation:

$x^{2} -(-p)x+(\frac{-1}{p})=0$

$x^{2} +px-\frac{1}{p}=0$

On taking LCM as $p$

$px^{2} +p^{2}x-1=0$

The quadratic polynomial is​ $px^{2} +p^{2}x-1=0$.