Name: Transformation of Roots of a Quadratic Equation
Uploaded: 2019-07-19
Duration: 11 min 45 s
Description: Here an equation and its roots are given, and we have to find new quadratic equations with new roots, which are simple multiples or factors of the original r...

Question

Solve

\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

Accepted Answer

Given $\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

We have to find the solution of given quadratic equation.

$∵ \frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

$\Rightarrow \frac{(x - 1) - (x - 2)}{(x - 1) (x - 2)} + \frac{(x - 2) - (x - 3)}{(x - 2) (x - 3)} + \frac{(x - 3) - (x - 4)}{(x - 3) (x - 4)} = \frac{1}{6}$

$\Rightarrow \frac{1}{(x - 2)} - \frac{1}{(x - 1)} + \frac{1}{(x - 3)} - \frac{1}{(x - 2)} + \frac{1}{(x - 4)} - \frac{1}{(x - 3)} = \frac{1}{6}$

$\Rightarrow \frac{1}{x - 4} - \frac{1}{x - 1} = \frac{1}{6}$ $\Rightarrow \frac{x - 1 - x + 4}{(x - 4) (x - 1)} = \frac{1}{6}$

$\Rightarrow (x - 4) (x - 1) = 18$ $\Rightarrow x^{2} - 5 x + 4 - 18 = 0$

$\Rightarrow x^{2} - 5 x - 14 = 0$ $\Rightarrow x^{2} - 7 x + 2 x - 14 = 0$

$\Rightarrow x (x - 7) + 2 (x - 7) = 0$ $\Rightarrow (x - 7) (x + 2) = 0$

Now if $x - 7 = 0$ then $x = 7$ and

if $x + 2 = 0$ then $x = - 2$

$∴$ The required solution is $x = - 2, 7$

Solve $\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

Important Questions on Quadratic Equation in One Variable

Learn and Prepare for any exam you want !

Solve 1(x-1)(x-2)+1(x-2)(x-3)+1(x-3)(x-4)=16

Important Questions on Quadratic Equation in One Variable

Learn and Prepare for any exam you want !

Solve $\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$