Question
Form a quadratic polynomial whose zeroes are $\frac{3-\sqrt{3}}{5}$ and $\frac{3+\sqrt{3}}{5}$
✓
Answer
$\frac{3-\sqrt{3}}{5}$ and $\frac{3+\sqrt{3}}{5}$ are the zeroes of the quadratic polynomial.
Let $\alpha=\frac{3-\sqrt{3}}{5}$ and $\beta=\frac{3+\sqrt{3}}{5}$
Sum of zeroes $=\alpha+\beta$
$\Rightarrow \frac{3-\sqrt{3}}{5}+\frac{3+\sqrt{3}}{5}=\frac{3-\sqrt{3}+3+\sqrt{3}}{5}=\frac{6}{5}$
Product of zeroes $=\alpha \beta$
$\Rightarrow\left(\frac{3-\sqrt{3}}{5}\right)\left(\frac{3+\sqrt{3}}{5}\right)=\frac{9-3}{25}=\frac{6}{25}$
The quadratic equation whose zeroes are $\alpha$ and $\beta$ are:
$\Rightarrow x^2-(\alpha+\beta) x+\alpha \beta=0$
$\Rightarrow x^2-\left(\frac{6}{5}\right) x+\left(\frac{6}{25}\right)=0$
Multiply both sides by
$25 x^2-30 x+6=0$
Therefore, the required polynomial is
$25 x^2-30 x+6=0$
Let $\alpha=\frac{3-\sqrt{3}}{5}$ and $\beta=\frac{3+\sqrt{3}}{5}$
Sum of zeroes $=\alpha+\beta$
$\Rightarrow \frac{3-\sqrt{3}}{5}+\frac{3+\sqrt{3}}{5}=\frac{3-\sqrt{3}+3+\sqrt{3}}{5}=\frac{6}{5}$
Product of zeroes $=\alpha \beta$
$\Rightarrow\left(\frac{3-\sqrt{3}}{5}\right)\left(\frac{3+\sqrt{3}}{5}\right)=\frac{9-3}{25}=\frac{6}{25}$
The quadratic equation whose zeroes are $\alpha$ and $\beta$ are:
$\Rightarrow x^2-(\alpha+\beta) x+\alpha \beta=0$
$\Rightarrow x^2-\left(\frac{6}{5}\right) x+\left(\frac{6}{25}\right)=0$
Multiply both sides by
$25 x^2-30 x+6=0$
Therefore, the required polynomial is
$25 x^2-30 x+6=0$
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