Question

find a quadratic polynomial each with given numbers as sum and product of its zeroes respectively -14,14​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★Sum of the zeroes is -14 of a quadratic polynomial

★Product of the zeroes is 14 of a Quadratic polynomial.

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★The Quadratic polynomial.

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We would use the formula for writing the Quadratic polynomial if the sum and the product of its zeroes is given

$\huge\tt{\red{\underline{Answer:}}}$

Let

• $\alpha$ as the first zero.
• $\beta$ as the second zero.

Now, by question,

$\green{\boxed{\alpha+\beta=(-14)}}$

$\green{\boxed{\alpha\beta=14}}$

______________________________________

We know when the sum and product of zeroes of a quadratic polynomial is given, then it can be written as, [say polynomial is p(x) ]

$\large\purple{\boxed{p(x) = x^{2}-x(\alpha+\beta) +(\alpha\beta) }}$

On substituting the values,

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

$\implies p(x) = x^{2}-x(-14) +14$

$\implies p(x) = x^{2}+14x+14$

${\underline{\boxed{x^{2}+14x+14}}}$

Therefore the required quadratic polynomial is $x^{2}+14x+14$.

Answer 2

Step-by-step explanation:

$\alpha + \beta = ( - 14)$

$\alpha \beta = 14$

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

$p(x) = x ^{2} - x( - 14) + 14$

$p(x) = x ^{2} + 14x + 14$