91Ó°ÊÓ

The substitution method simplifies complex equations by introducing a new variable. In our example, we start with the equation:
$$2 x^{-2 / 5}-x^{-1 / 5}-1=0.$$
This is a complicated equation due to the exponents on x. By letting y be equal to one of the terms, we can turn the equation into a more familiar form. Here, we make the substitution:
Let y = x^{-1/5}.
Now, the equation transforms to:
$$2y^2 - y - 1 = 0.$$
This substitution helps to manage the tricky exponents and turns our problem into a standard quadratic equation, which is much easier to solve.

Once the equation is simplified using substitution, we can solve it using the quadratic formula. For a quadratic equation in the form ax^2 + bx + c = 0, the quadratic formula is:
$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
Here, the coefficients are

• a = 2
,
• b = -1
,
• c = -1
.

We substitute these values into the quadratic formula:
$$y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm\sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}.$$ This gives us two solutions for y:

• y = 1
and
• y = -1/2

Now the quadratic equation is solved, and we can return to our original variable.

Finally, we need to revert back to our original variable and understand the nature of the solutions. Recall the substitution:
y = x^{-1/5}.
Our two solutions for y were 1 and -1/2. We convert back to x for each solution.

For x^{-1/5} = 1, raising both sides to the power of -5 gives us:
$$x = 1^{-5} = 1.$$
For x^{-1/5} = -1/2, raising both sides to the power of -5 would result in a non-real number because we are raising a negative number to a negative fractional power. In this context, such solutions are considered non-real and therefore are discarded.

Real solutions make sense within the original exercise while non-real solutions do not. Understanding this distinction is crucial in algebraic equations.