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How do you solve #15x^(-2)+7x^(-1)-2=0#?

Answer 1

Hazel Anglin

Multiply through by #x^2# to turn it into a standard quadratic equation, and then solve with the quadratic formula

First, we multiply each side of the equation by #x^2#.

#x^2(15x^-2 + 7x^-1 - 2) = 0x^2# #=>15 + 7x - 2x^2 = 0# (by applying the rule #x^a * x^b = x^(a+b)#)

Rearranging the terms gives us a familiar quadratic form.

#-2x^2 + 7x + 15 = 0#

Finally, we apply the quadratic formula #x = (-b +- sqrt(b^2 -4ac))/(2a)# where #a = -2#, #b = 7#, and #c = 15#

This gives

#x = -3/2# or #x = 5#.

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Answer 2

Savannah Anderson

To solve the equation 15x^(-2) + 7x^(-1) - 2 = 0:

Multiply the entire equation by x^2 to clear the denominators.
Rearrange the equation to standard form.
Use the quadratic formula to solve for x.
Substitute the values of a, b, and c into the quadratic formula and simplify.

The solutions will be the values of x that make the equation true.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.