Given that a and b are the roots of the equation $$x^2\ -\ 13x\ +\ 42\ =\ 0$$. Which of the following is the equation with $$\dfrac{1}{a}$$ and $$\dfrac{1}{b}$$ as its roots?
Solution
Given that a and b are the roots of the equation $$x^2\ -\ 13x\ +\ 42\ =\ 0$$
Sum of the roots = a + b = $$-\dfrac{b}{a}$$ = $$-\dfrac{\left(-13\right)}{1}$$ = 13
Product of the roots = ab = $$\dfrac{c}{a}$$ = $$\dfrac{42}{1}$$ = 42
Let the new quadratic equation be $$x^2\ +\ cx\ +\ d\ =\ 0$$. For the new equation with roots as $$\dfrac{1}{a}$$ and $$\dfrac{1}{b}$$, the sum of the roots and product of the roots can be calculated as
Sum of the roots = $$\dfrac{1}{a}\ +\ \dfrac{1}{b}$$ = $$\dfrac{a\ +\ b}{ab}$$ = $$\dfrac{13}{42}$$ = -c
Product of the roots = $$\dfrac{1}{a}\times\ \dfrac{1}{b}$$ = $$\dfrac{1}{ab}$$ = $$\dfrac{1}{42}$$ = d
So, the new equation becomes,
$$x^2\ -\ \dfrac{13}{42}x\ +\ \dfrac{1}{42}\ =\ 0$$
Multiplying the whole equation by 42, we get,
$$42x^2\ -\ 13x\ +\ 1\ =\ 0$$
The correct answer is option A.
