Question

solve the equation after making an appropriate substitution.
$35a^{-2} + 44a^{-1} - 7 = 0$

the solution set is
(simplify your answer type an exact answer, using radicals as needed. express complex numbers in terms of $i$. use a comma to separate answers as needed. type each solution only once.)

Explanation:

Step1: Substitute \( x = a^{-1} \)

Let \( x = a^{-1} \), then \( a^{-2} = x^2 \). The equation \( 35a^{-2} + 44a^{-1} - 7 = 0 \) becomes \( 35x^2 + 44x - 7 = 0 \).

Step2: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( ax^2 + bx + c = 0 \). Here, \( a = 35 \), \( b = 44 \), \( c = -7 \).
First, calculate the discriminant \( \Delta = b^2 - 4ac = 44^2 - 4\times35\times(-7) = 1936 + 980 = 2916 \).
Then, \( x = \frac{-44 \pm \sqrt{2916}}{2\times35} = \frac{-44 \pm 54}{70} \).
We get two solutions for \( x \):

• \( x_1 = \frac{-44 + 54}{70} = \frac{10}{70} = \frac{1}{7} \)
• \( x_2 = \frac{-44 - 54}{70} = \frac{-98}{70} = -\frac{7}{5} \)

Step3: Substitute back to find \( a \)

Since \( x = a^{-1} = \frac{1}{a} \), then:

• For \( x = \frac{1}{7} \), \( \frac{1}{a} = \frac{1}{7} \), so \( a = 7 \).
• For \( x = -\frac{7}{5} \), \( \frac{1}{a} = -\frac{7}{5} \), so \( a = -\frac{5}{7} \).

Answer:

\( 7, -\frac{5}{7} \)