Question

If ${x^2} - 8x - 1 = 0$ then prove ${x^2} + \dfrac{1}{{{x^2}}}$

Answer 1

Hint: The given equation is a quadratic equation since the highest power of x is 2. We should simplify the equation and try to find out the value of the equation ${x^2} + \dfrac{1}{{{x^2}}}$ using the quadratic equation given. We can either find the answer in terms of x or find the numerical value of the equation (either by
simplification or finding the value of x).

Complete step-by-step answer:
The given quadratic equation is ${x^2} - 8x - 1 = 0$.
When we try to simplify the equation, take the term $8x + 1$to the other side. The equation becomes:
${x^2} = 8x + 1$
To find the value of ${x^2} + \dfrac{1}{{{x^2}}}$ we need ${x^2}$and its reciprocal $\dfrac{1}{{{x^2}}}$.
We know that:
${x^2} = 8x + 1$ and $\dfrac{1}{{{x^2}}} = \dfrac{1}{{8x + 1}}$
Now when we add the two values, we get:
$(8x + 1) + \dfrac{1}{{(8x + 1)}} = \dfrac{{{{(8x + 1)}^2} + 1}}{{(8x + 1)}}$
By further simplification we get:
$\dfrac{{64{x^2} + 1 + 16x + 1}}{{8x + 1}}$(${(8x + 1)^2} = 64{x^2} + 16x + 1$)
$\dfrac{{64{x^2} + 16x + 2}}{{8x + 1}}$
The final answer is ${x^2} + \dfrac{1}{{{x^2}}} = \dfrac{{64{x^2} + 16x + 2}}{{8x + 1}}$
This is the answer in terms of x. We can also find the numerical value by the following simplification:
${x^2} - 8x - 1 = 0$
By dividing the entire equation with x, we get:
$x - 8 - \dfrac{1}{x} = 0$
By further simplification we get:
$x - \dfrac{1}{x} = 8$
By squaring on both the sides we get:
${(x - \dfrac{1}{x})^2} = {(8)^2}$
${x^2} - 2(x)(\dfrac{1}{x}) + \dfrac{1}{{{x^2}}} = 64$
${x^2} + \dfrac{1}{{{x^2}}} = 64 + 2 = 66$
Hence the numerical value is 66.
${x^2} + \dfrac{1}{{{x^2}}} = \dfrac{{64{x^2} + 16x + 2}}{{8x + 1}} = 66$

Note: Roots of the equation can be found out by using discriminant method and formula or we can find it by splitting the middle term and simplifying. Depending upon the question, we need to change our simplification techniques. We need to be careful with the positive and negative signs while squaring and adding the equations.