Question

What is the solution set to the following equation?

x4+2x−48=0

Select one:
a. { ±2i2–√,±6–√ }
b. { ±i6–√,±2–√ }
c. { ±2i2–√,±3 }
d. { ±i3–√,±22–√ }

Answer 1

The solution set of the equation x^2 + 2x - 48 = 0 is x = -1 ± 7

How to determine the solution set of the equation?

The equation is given as:

x^2 + 2x - 48 = 0

A quadratic equation is represented as:

ax^2 + bx + c = 0

By comparing both equations, we have

a = 1, b = 2 and c = -48

The solution of the quadratic equation is then calculated using

x = (-b ± √(b^2 - 4ac))/2a

Substitute values for a, b and c in the above equation

x = (-2 ± √(2^2 - 4 * 1 * -48))/2 * 1

This gives

x = (-2 ± √196)/2

Evaluate the square root of 196

x = (-2 ± 14)/2

Divide through by 2

x = -1 ± 7

Hence, the solution set of the equation x^2 + 2x - 48 = 0 is x = -1 ± 7

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