MCQ
The quadratic equation whose roots are $7 + \sqrt{3}$ and $7 - \sqrt{3}$ is :
- ✓$ x^2-14 x+46=0 $
- B$ x^2-14 x-46=0 $
- C$ x^2+14 x+46=0 $
- D$ x^2+14 x-46=0 $
✓
Answer
Correct option: A.
$ x^2-14 x+46=0 $
Given : $\alpha = 7+\sqrt{3}$ and $\beta= 7-\sqrt{3}$
$\therefore\text{x}^{2}-(\alpha+\beta)\text{ x}+\alpha\beta = {0}$
$\Rightarrow\text{x}^{2} -(7+\sqrt{3}+7-\sqrt{3})\text{ x}+(7+\sqrt{3}) (7-\sqrt{3})=0$
$\Rightarrow\text{x}^{2} - {14}\text{x}+(49-3) = 0$
$\Rightarrow\text{x}^{2} - {14}\text{x}+46 = 0$
$\therefore\text{x}^{2}-(\alpha+\beta)\text{ x}+\alpha\beta = {0}$
$\Rightarrow\text{x}^{2} -(7+\sqrt{3}+7-\sqrt{3})\text{ x}+(7+\sqrt{3}) (7-\sqrt{3})=0$
$\Rightarrow\text{x}^{2} - {14}\text{x}+(49-3) = 0$
$\Rightarrow\text{x}^{2} - {14}\text{x}+46 = 0$
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