Question

.If x² – 16x – 59 = 0, then what is the value of (x – 6)² + [1/(x – 6)²]?​

Answer 1

Given :

The quadratic equation as x² – 16 x – 59 = 0

To Find :

The value of  ( x - 6 )² + $\dfrac{1}{(x-6)^{2} }$

Solution :

∵  quadratic equation as x² – 16 x – 59 = 0

So, For standard quadratic equation

i.e   a x² + b x + c = 0

Roots can be find as formula

  x = $\dfrac{- b\pm \sqrt{b^{2}-4\times a\times c}}{2 \times a}$

So, while comparing with given equation , we get

 x = $\dfrac{16\pm \sqrt{(-16)^{2}-4\times 1\times (-59)}}{2 \times 1}$

   = $\dfrac{16\pm \sqrt{(256)+236}}{2 }$

   = $\dfrac{16\pm 22.18}{2 }$

So, x = 19.09  , - 3.09

Now ,

For  x = 19.09

( x - 6 )² + $\dfrac{1}{(x-6)^{2} }$  =  ( 19.09 - 6 )² + $\dfrac{1}{(19.09-6)^{2} }$

                              = ( 13.09 )² + $\dfrac{1}{(13.09)^{2} }$

                              = 171.34 + 0.005836

                             = 171.3458

Again

For  x = - 3.09

( x - 6 )² + $\dfrac{1}{(x-6)^{2} }$  =  ( - 3.09 - 6 )² + $\dfrac{1}{(- 3.09-6)^{2} }$

                              = ( - 9.09 )² + $\dfrac{1}{(- 9.09)^{2} }$

                              = 82.62 + 0.0121

                             = 82.632

Hence, The value of ( x - 6 )² + $\dfrac{1}{(x-6)^{2} }$ is  171.3458  and   82.632  Answer