Question

The quadratic equations having the roots $\frac{1}{10-\sqrt{72}}$ and $\frac{1}{10+6\sqrt{2}}$ is

• A. $28x^2-20x+1=0$
• B. $20x^2-28x+1=0$
• C. $x^2-20x+28=0$
• D. $x^2-28x+20=0$

Answer 1

The correct option is A

$28x^2-20x+1=0$

Explanation for correct option:

Find the quadratic equation based on given roots

Let $\alpha =\frac{1}{10-\sqrt{72}}$

multiplying $10+\sqrt{72}$ to both numerator and denominator

$$\begin{gathered} \alpha =\frac{1}{10-\sqrt{72}}\times \frac{10+\sqrt{72}}{10+\sqrt{72}} \\ \alpha =\frac{10+\sqrt{72}}{100-72} \\ \alpha =\frac{10+6\sqrt{2}}{28} \end{gathered}$$

and $\beta =\frac{1}{10+6\sqrt{2}}$

multiplying $10-6\sqrt{2}$ to both numerator and denominator

$$\begin{gathered} \beta =\frac{1}{10+6\sqrt{2}}\times \frac{10-6\sqrt{2}}{10-6\sqrt{2}} \\ \beta =\frac{10-6\sqrt{2}}{100-72} \\ \beta =\frac{10-6\sqrt{2}}{28} \end{gathered}$$

a quadratic equation having roots $\alpha$ and $\beta$ is $x^2-(\alpha +\beta )x+\alpha ·\beta =0$

$$\begin{gathered} x^2-\left(\frac{10+6\sqrt{2}}{28}+\frac{10-6\sqrt{22}}{28}\right)x+\left(\frac{10+6\sqrt{2}}{28}\right)\left(\frac{10-6\sqrt{2}}{28}\right)=0 \\ {x}^2-\left(\frac{20}{28}\right)x+\frac{100-72}{28\times 28}=0 \\ 28x^2-20x+\frac{28}{28}=0 \\ 28x^2-20x+1=0 \end{gathered}$$

Hence, $\frac{1}{10-\sqrt{72}}$ and $\frac{1}{10+6\sqrt{2}}$ are roots of $28x^2-20x+1=0$, so, option (A) is the correct answer.