Solve the equation after making an appropriate substitution.
5x^(-2)-29x^(-1)-6=0
Determine the appropriate substitution using the new variable “u.”
The correct answer and explanation is :
The given equation is:
[
5x^{-2} – 29x^{-1} – 6 = 0
]
Step 1: Make a substitution.
We begin by observing that the exponents of (x) are negative powers. A standard approach in solving equations with negative exponents is to use a substitution to simplify the equation.
Let’s define a new variable ( u ) such that:
[
u = x^{-1}
]
This implies that:
[
x^{-2} = u^2
]
Step 2: Substitute the values.
Now, substitute ( u = x^{-1} ) and ( x^{-2} = u^2 ) into the original equation. This gives:
[
5u^2 – 29u – 6 = 0
]
Step 3: Solve the quadratic equation.
Now, we have a quadratic equation in terms of (u):
[
5u^2 – 29u – 6 = 0
]
We can solve this quadratic equation using the quadratic formula. The quadratic formula is given by:
[
u = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
]
For the equation ( 5u^2 – 29u – 6 = 0 ), the coefficients are:
- (a = 5)
- (b = -29)
- (c = -6)
Substitute these values into the quadratic formula:
[
u = \frac{-(-29) \pm \sqrt{(-29)^2 – 4(5)(-6)}}{2(5)}
]
[
u = \frac{29 \pm \sqrt{841 + 120}}{10}
]
[
u = \frac{29 \pm \sqrt{961}}{10}
]
[
u = \frac{29 \pm 31}{10}
]
Step 4: Calculate the two solutions for (u).
Now, we calculate the two possible values for (u):
- (u = \frac{29 + 31}{10} = \frac{60}{10} = 6)
- (u = \frac{29 – 31}{10} = \frac{-2}{10} = -\frac{1}{5})
Step 5: Solve for (x).
Recall that (u = x^{-1}), so:
- If (u = 6), then (x^{-1} = 6), which gives (x = \frac{1}{6}).
- If (u = -\frac{1}{5}), then (x^{-1} = -\frac{1}{5}), which gives (x = -5).
Final Answer:
Thus, the two solutions to the original equation are:
[
x = \frac{1}{6} \quad \text{or} \quad x = -5
]
Explanation:
In solving the equation, we first simplified it by making the substitution ( u = x^{-1} ), which turned the original equation into a quadratic equation in terms of (u). This allowed us to apply the quadratic formula to find two possible values for (u). Once the values for (u) were found, we simply solved for (x) by taking the reciprocal of each value of (u), leading to the two solutions for (x). This substitution method reduces the complexity of the equation and makes it easier to solve.