Solve the equation: \$(2/x) + (3/(x^2)) = 5\$.

The given equation

\$(2/x) + (3/(x^2)) = 5\$

To solve this equation, let’s first make a substitution. Let’s set a variable, let’s say y, equal to \$x^(-1)\$. So,

\$ y = 1/x \$

Now we can rewrite the equation in terms of y, rearranging the equation, we get:

\$ 2y + 3y^2 = 5 \$

\$ 3y^2 + 2y - 5 = 0 \$

Now, we have a standard quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula

\$ x = (-b ± \sqrt(b^2 - 4ac)) / (2a) \$

In this case, a = 3, b = 2, and c = - 5. Plugging these values into the quadratic formula, we have:

\$ y = (-2 ± sqrt(2^2 - 4 * 3 * -5)) / (2 * 3) \$

\$ y = (-2 ± 8) / 6 \$

So, we have two possible solutions for y then we get x value

\$ y_1 = (8 - 2) / 6 = 6/6 = 1 \$ and

\$ y_2 = (-8 - 2) / 6 = -10/6 = - 5/3 \$

\$ 1/x = 1 and 1/x = - 5/3 \$

\$ x_1 = 1/1 and x_2 = 1/(- 5/3) \$

\$ x_1 = and x_2 = - 3/5 \$

Therefore, the solutions to the original equation are \$ x_1 = 1 and x_2 = - 3/5 \$.

This is the incorrect choice. The calculation showed that after rearranging and applying the quadratic formula, these values did not satisfy the equation

\$ 1, - 5/3\$

This choice is correct because after applying the quadratic formula or another method, these were the values of x that satisfied the original equation

\$ 1, - 3/5\$

Upon solving the equation, we find that x = −1 does not satisfy the given equation, making this option incorrect

1, -1

In this case, since option B correctly identifies the solutions, option D would not be the right choice

None of these above

Option

B

Can you summarize what you’ve understood in the above steps?