Question

Question 2
A quadratic polynomial, who zeroes are -3 and 4, is
(a) $x^2–x+12$
(b) $x^2+x+12$
(c) $\frac{x^2}{2}-\frac{x}{2}-6$
(d) $2x^2+2x–24$

Answer 1

Let $a{x}^2+bx+c$ be a required polynomial whose zeroes are - 3 and 4
Then, sum of zeroes = -3 + 4 = 1 $[sumofzeores=\frac{-b}{a}]$
$\Rightarrow \frac{-b}{a}=\frac{1}{1}\Rightarrow \frac{-b}{a}=-\frac{(-1)}{1}$
and product of zeroes = $-3\times 4=-12$ $[productofzeroes=\frac{c}{a}]$
$\Rightarrow \frac{c}{a}=\frac{-12}{1}$
From above we can conclude that
a = 1, b = - 1 and c = -12
$\therefore$ required polynomial = $1x^2–1x–12$
$=x^2–x-12$
$=\frac{x^2}{2}-\frac{x}{2}-6$
We know that, if we multiply / divide any polynomial by any constant, then the zeroes of polynomial do not change.

$Alternatemethod$
Let the zeroes of a quadratic polynomial are $\alpha =-3and\beta =4$
Then, sum of zeroes = $\alpha +\beta =-3+4=1$
and product of zeroes =$\alpha \beta =(-3)\left(4\right)=-12$
$\therefore$ Required polynomial = $x^2$ - (sum of zeroes)$x$ + (product of zeroes)
$=x^2–\left(1\right)x+(-12)=x^2–x–12$
$=\frac{x^2}{2}-\frac{x}{2}-6$