Question

3. The quadratic polynomial, the sum of whose zeros is -5and their product is 6,is​

Answer 1

Step-by-step explanation:

I hope this will help u .

Answer 2

Given:-

• Sum of the zeroes = -5
• Product of the zeroes = 6

To Find:-

The Quadratic Equation.

Assumption:-

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

Therefore,

Solution:-

We have $\alpha$ and $\beta$ as the two zeroes of the polynomial.

Therefore,

ATQ,

Sum of zeroes = -5

$\sf{\implies \alpha + \beta = -5}$

And

Product of zeroes = 6

$\sf{\implies \alpha\beta = 6}$

Now,

We know, A quadratic equation is always in the form:-

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

Therefore, Substituting the values,

$\sf{x^2 - (-5)x + 6\:\:[\because \alpha + \beta = -5\:\:and\:\: \alpha\beta = 6]}$

$\sf{\therefore The\:Quadratic\:Equation\:is\:\underline{ x^2 + 5x + 6}}$

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Verification:-

Now let us verify whether our equation is correct or not.

We have quadratic equation = x² + 5x + 6

Let us split the middle term,

$\sf{x^2 + 3x + 2x + 6}$

= $\sf{x(x + 3) + 2(x + 3)}$

= $\sf{(x+3)(x+2)}$

Either,

$\sf{x+3 = 0}$

= $\sf{x = -3}$

Or,

$\sf{x+2 = 0}$

= $\sf{x = -2}$

Therefore the two zeroes of the equation are -3 and -2

Now,

We know,

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

= $\sf{-3-2 = \dfrac{-(5)}{1}}$

= $\sf{-5 = -5}$

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

= $\sf{-2\times -3 = \dfrac{6}{1}}$

= $\sf{6 = 6}$

Hence Verified!!

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