Question 16

If $$x^2 - 16x - 59 = 0$$, then what is the value of $$(x - 6)^2 + \left[\frac{1}{(x - 6)^2}\right]$$?

Solution

$$x^2-16x-59=0$$

So, $$x=\frac{16\pm\sqrt{256+236}}{2}=19\ or\ -3\ .$$

when , x=19 :

$$\left(x-6\right)^2+\frac{`1}{\left(x-6\right)^2}=169\ .$$

Or, when x=-3 :

$$\left(x-6\right)^2+\frac{`1}{\left(x-6\right)^2}=81.01\ .$$

So, B is correct choice.

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