Substitute y = x 2 to transform the equation into a quadratic equation in y .
Factor the quadratic equation ( y − 7 ) ( y + 2 ) = 0 and solve for y , obtaining y = 7 and y = − 2 .
Substitute back x 2 for y to get x 2 = 7 and x 2 = − 2 .
Solve for x to find the solutions x = ± 7 and x = ± i 2 . The final answer is x = ± 7 and x = ± i 2 .
Explanation
Problem Analysis We are given the equation x 4 − 5 x 2 − 14 = 0 and asked to find its solutions by factoring.
Substitution Let's make a substitution to simplify the equation. Let y = x 2 . Then the equation becomes y 2 − 5 y − 14 = 0 .
Factoring the Quadratic Now we factor the quadratic equation in y . We are looking for two numbers that multiply to -14 and add to -5. These numbers are -7 and 2. So, we can factor the equation as ( y − 7 ) ( y + 2 ) = 0 .
Solving for y Solving for y , we have two possible values: y − 7 = 0 or y + 2 = 0 . This gives us y = 7 or y = − 2 .
Substituting Back Now we substitute back x 2 for y . We have x 2 = 7 or x 2 = − 2 .
Solving for x Solving for x in each case:
For x 2 = 7 , we have x = ± 7 .
For x 2 = − 2 , we have x = ± − 2 = ± i 2 .
Final Answer Therefore, the solutions are x = ± 7 and x = ± i 2 .
Examples
Understanding polynomial equations is crucial in many fields, such as physics and engineering. For example, when analyzing the motion of a projectile, you might encounter an equation similar to the one we solved. Factoring helps simplify these equations, allowing you to find key parameters like the launch angle or initial velocity needed to hit a target. This skill is also valuable in economics for modeling market behavior and predicting trends.
The solutions of the equation x 4 − 5 x 2 − 14 = 0 are x = ± 7 and x = ± i 2 . Therefore, the correct answer is option D. We factored the equation using a substitution that simplified it into a quadratic form, allowing us to find the solutions easily.
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