Question

find a quadratic polynomial with given numbers as the sum and product of zeroes respectively -2and-3​

Answer 1

Answer:

x^2-(sum of zeroes) x+product of zeroes

x^2 -(-2) x+(-3)

x^2+2x-3

Answer 2

Given:-

• Sum of zeroes = -2
• Product of zeroes = -3

To Find:-

The Quadratic polynomial.

Assumption:-

Let $\alpha$ and $\beta$ be the two zeroes of the polynomial.

Solution:-

ATQ,

Sum of zeroes = -2

$\alpha + \beta$ = -2

Product of zeroes = -3

$\alpha\beta$ = -3

We know,

A quadratic equation is always in the form of:-

$\underline{\boxed{\bf{x^2 - (\alpha + \beta)x + \alpha\beta}}}$

Therefore,

Substituting the values:-

$\sf{x^2 - (-2)x + (-3)}$

= $\sf{x^2 + 2x - 3}$

Therefore the required quadratic polynomial is $\underline{\sf{x^2 + 2x -3}}$

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Verification:-

Now let us verify whether the quadratic equation we got is correct or not.

Let us first find the zeroes of x² + 2x - 3.

By splitting the middle term,

= $\sf{x^2 + 3x - x - 3}$

= $\sf{x(x+3)-1(x+3)}$

= $\sf{(x+3)(x-1)}$

Either,

$\sf{x+3 = 0}$

= $\sf{x = -3}$

Or,

$\sf{x-1 = 0}$

= $\sf{x = 1}$

Therefore the two zeroes of the polynomial x² + 2x - 3 are -3 and 1

Now let us verify the relation between their coefficients.

We know,

Sum of zeroes = $\sf{\dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

And

Product of zeroes = $\sf{\dfrac{Constant\:Term}{Coefficient\:of\:x^2}}$

Therefore substituting the values,

For sum of zeroes,

$\sf{-3+1 = \dfrac{-2}{1}}$

= $\sf{-2 = -2}$

For product of zeroes,

$\sf{-3\times1 = \dfrac{-3}{1}}$

= $\sf{-3 = -3}$

Hence Verified!!!

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