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