Question

The sum and product of zeroes of a quadratic polynomial in x are -6and-7 respectively write quadratic polynomial

Answer 1

In the above Question , the following information is given -

The sum and product of zeroes of a quadratic polynomial in x are -6 and -7 .

To find -

Find the respective quadratic polynomial .

Solution -

A quadratic polynomial can be written as -

=> x² - ( Sum of zeroes ) x + ( Product of zeroes )

Now , we have -

Sum of zeroes = -6

Product of zeroes = -7 .

Substituting these , the new quadratic equation becomes -

=> x² - ( -6 x ) - 7

=> x² + 6x - 7 .

Verification -

Given polynomial -

=> x² + 6x - 7

=> x² + 7x - x - 7

=> x ( x + 7 ) - 1 ( x + 7 )

=> ( x - 1 )( x + 7 )

=> The required roots are 1 and - 7

Sum of zeroes => -6

Product of zeroes => -7 .

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Answer 2

ANSWER:-

➣

• Given:-

Sum of zeroes of Quadratic equation in x =$\bf{-6}$

Product of zeroes of Quadratic equation in x =$\bf{-7}$

• To Find:-

Quadratic Polynomial of the zeroes =$\bf{?}$

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• Solution:-

We know:-

$\implies{x^2 - (Sum \: of \: zeroes)x + (Product \: of \: zeroes)}$

• Given that:-

• Sum of zeroes =$\bf{-6}$

• Product of zeroes =$\bf{-7}$

Substituting Values in the above Equation:-

$\implies x^2 - (-6x) - 7$

$\implies{\bf{ x^2 + 6x - 7}}$

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➣ Verification:-

$\longrightarrow{x^2 + 6x - 7}$

• Splitting middle term:-

$\longrightarrow{x^2 + 7x - x - 7}$

$\longrightarrow{x(x+7) - 1(x+7)}$

$\longrightarrow{(x-1)(x+7)}$

Roots:-

$\implies(x-1) =0$

$\bf{x = 1}$

$\\ \implies(x+7) =0$

$\bf{ x = -7 }$

Hence, the required Roots = $\bf{1 \: and \: -7}$

Therefore,

• Sum of zeroes = $\bf{-6}$

• Product of zeroes = $\bf{-7}$

Hence, Verified ✅

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Some Useful Information:-

• The sum of zeros is equal to the negative of the coefficient of $x$ by coefficient of $x^2$.

• The product of zeros is equal to constant term by coefficient of $x^2$

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