Question

from a quadratic polynomial whose one of the zeroes is +15 and sum of zeroes is 42​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Correct \ Question \colon}}}}$

Form a quadratic polynomial whose sum of zeros is 42 and one of the zero is 15.

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Given,

• The sum of zeros is 42

• One of the zero is 15

Let the unknown zero be $\alpha$

Implies,

$\large{ \sf{15 + \alpha = 42}} \\ \\ \large{ \leadsto \: \sf{ \alpha = 42 - 15}} \\ \\ \huge{ \leadsto \: \boxed{ \boxed{\tt{ \alpha = 27}}}}$

Product of Zeros

$\large {\alpha \beta = (15)(27)} \\ \\ \huge{ \rightarrow \: \tt{ \alpha \beta = 405}}$

Required Polynomial

$\sf{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta } \\ \\ = \huge{ \sf{x {}^{2} - 42x + 405 }}$

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Answer 2

$\bold{\Huge{\underline{\boxed{\sf{\purple{ANSWER\::}}}}}}$

$\bold{\Large{\underline{Given\::}}}}}$

In a quadratic polynomial whose one of the zeroes is 15 & sum of the zeroes is 42.

$\bold{\Large{\underline{To\:find\::}}}}}$

The required quadratic polynomial.

$\bold{\Large{\underline{Explanation\::}}}}}$

We know that general form of polynomial= Ax² + Bx + C

• Sum of the zeroes:

→ $\bold{\alpha +\beta =\frac{-b}{a} }$

A/q

→ $\bold{15+\beta =42}$

→ $\bold{\beta =\:42\:-\:15}$

→ $\bold{\beta\:=\:27}$

• Product of the zeroes:

→ $\bold{\alpha \beta =\frac{c}{a} }$

Therefore,

$\bold{we\:have\begin{cases}\sf{\alpha =15}\\ \sf{\beta =27}\end{cases}}$

→ $\bold{\alpha \beta =(15*27)}$

→ $\bold{\alpha \beta =405}$

Thus,

The required quadratic polynomial is x² - (α+β)x + αβ

→ x² - 42x + 405