MCQ
A quadratic polynomial with zeroes $\frac{1}{4}$ and $-1$ is :
- A$ 4 x^2+3 x+1 $
- B$ 4 x^2-3 x+1 $
- C$ 4 x^2-3 x-1 $
- ✓$ 4 x^2+3 x-1 $
✓
Answer
Correct option: D.
$ 4 x^2+3 x-1 $
$\text{x}^{2} -(\alpha+\beta)\text{ x}+(\alpha+\beta) = {0}$
Here $\alpha+\beta = \frac{1}{4} = \frac{1-4}{4} = \frac{-3}{4}$ and $\alpha\beta$
$ =\frac{1}{4}\times(-1) = \frac{-1}{4} = \frac{\text{c}}{\text{a}}$
$\text{x}^{2}+\frac{3}{4} \text{ x}-\frac{1}{4} ={0}$
$\frac{\text{x}^{2}+{3}\text{x}-{1}}{4} =0 \ ($By $\text{L.C.M)}$
$4\text{x}^{2}+3\text{x}-1 = {0}$
Here $\alpha+\beta = \frac{1}{4} = \frac{1-4}{4} = \frac{-3}{4}$ and $\alpha\beta$
$ =\frac{1}{4}\times(-1) = \frac{-1}{4} = \frac{\text{c}}{\text{a}}$
$\text{x}^{2}+\frac{3}{4} \text{ x}-\frac{1}{4} ={0}$
$\frac{\text{x}^{2}+{3}\text{x}-{1}}{4} =0 \ ($By $\text{L.C.M)}$
$4\text{x}^{2}+3\text{x}-1 = {0}$
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