Question

The equation whose roots are reciprocal of the roots of the equation $3{{x}^{2}}-20x+17=0$ is
( a ) $3{{x}^{2}}+20x-17=0$
( b ) $17{{x}^{2}}-20x+3=0$
( c ) $17{{x}^{2}}+20x+3=0$
( d ) none of these

Answer 1

Hint: In this question, we are given a quadratic equation and we have to find the equation whose roots are reciprocal to the roots of the given quadratic equation. For this, first we find the roots of the given quadratic equation by putting the values in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Then by doing the reciprocal of roots and finding the sum and the product of that roots, we are able to find the quadratic equation.

Formula used:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step by step Solution:
Here we are given the equation $3{{x}^{2}}-20x+17=0$
Compare the above equation with the standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
a = 3, b = -20, c = 17
We can the formula to find the quadratic equation is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By putting the values of a, b, and c in the above equation, we get
$x=\dfrac{-(-20)\pm \sqrt{{{(-20)}^{2}}-4(3)(17)}}{2(3)}$
By simplifying the above equation, we get
$x=\dfrac{-(-20)\pm \sqrt{196}}{6}$
That is $x=\dfrac{20\pm 14}{6}$
Then $ x = \dfrac{20+14}{6},\dfrac{20-14}{6}$
$x = \dfrac{34}{6},\dfrac{6}{6}$
Then $x = \dfrac{17}{3},1$
Hence the roots of the given equation are $\dfrac{17}{3},1$
So we need to form an equation whose roots are $\dfrac{3}{17},1$
Sum of roots = $\dfrac{3}{17}+1$= $\dfrac{20}{17}$
And the product of roots = $\dfrac{3}{17}\times 1$= $\dfrac{3}{17}$
We know in a quadratic equation, there is $ x^2$ -(sum of roots)$x$ +product of roots
Now By putting the values of the sum of roots and the product of roots, we get
Hence the required equation is ${{x}^{2}}-\dfrac{20}{17}x+\dfrac{3}{17}=0$
That is $17{{x}^{2}}-20x+3=0$

Therefore, the correct option is (b).

Note:To find the roots of the quadratic equation we use the formula:-
As we know standard form of quadratic equation is $a{{x}^{2}}+bx+c=0$
Then its roots are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$