MCQ
A quadratic polynomial whose zeroes are $-3$ and $6, $ is :
- A$\text{x}^{2}-{3}\text{x}+{18}$
- B$\text{x}^{2}+{3}\text{x}+{18}$
- ✓$\frac{\text{x}^{2}}{6}-\frac{\text{x}}{2}-{3}$
- D$\text{x}^{2}+{3}\text{x}-{18}$
✓
Answer
Correct option: C.
$\frac{\text{x}^{2}}{6}-\frac{\text{x}}{2}-{3}$
Here $\alpha+\beta = -{3}+{6} = \frac{3}{1}=\frac{-(-3)}{1} = \frac{\text{b}}{\text{a}}$
And $\alpha\beta = (-3)\times{6} = \frac{-18}{1}=\frac{\text{c}}{\text{a}}$
on comparing we get $a = 1, b = -3, c = -18$
putting these values in the general from of quadratic polynomial
$\text{ax}^{2}+\text{bx}+\text{c} = \frac{\text{x}^{2}-{3}\text{x}-{18}}{6} $
$= \frac{\text{x}^{2}}{6} -\frac{\text{x}}{2}-{3}\ [$Dividing all terms by $6]$
And $\alpha\beta = (-3)\times{6} = \frac{-18}{1}=\frac{\text{c}}{\text{a}}$
on comparing we get $a = 1, b = -3, c = -18$
putting these values in the general from of quadratic polynomial
$\text{ax}^{2}+\text{bx}+\text{c} = \frac{\text{x}^{2}-{3}\text{x}-{18}}{6} $
$= \frac{\text{x}^{2}}{6} -\frac{\text{x}}{2}-{3}\ [$Dividing all terms by $6]$
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