Question

Form a quadratic equation whose roots are the numbers $\frac{1}{10-\sqrt{72}}$ and $\frac{1}{10+6\sqrt{2}}$.

Answer 1

The standard form of quadratic equation is $a{x}^2+bx+c=0$

it is written in the form of $x^2+\left(sumofroots\right)x+\left(productofroots\right)$

sum of roots =$\frac{1}{10-\sqrt{72}}+\frac{1}{10+6\sqrt{2}}$

=$\frac{10+6\sqrt{2}+10-6\sqrt{2}}{(10{)}^2-(6\sqrt{2}{)}^2}$

=$\frac{20}{28}$

and the product of root is $\frac{1}{10-6\sqrt{2}}$$\times \frac{1}{10+6\sqrt{2}}$=$\frac{1}{28}$

put the value value on the equation $x^2-\frac{20}{28}x$+$\frac{1}{28}=0$

$28x^2+20x+1=0$ this is the standard quadratic equation.