$x^2+5 x+\frac{36}{x^2+5 x}=0$
Answer (1)
Substitute y = x 2 + 5 x to simplify the equation to y + y 36 = 0 .
Solve for y to get y = ± 6 i .
Substitute back to get two quadratic equations: x 2 + 5 x − 6 i = 0 and x 2 + 5 x + 6 i = 0 .
Solve the quadratic equations using the quadratic formula to find the four complex solutions: x ≈ 0.231 ± 1.099 i , − 5.231 ∓ 1.099 i .
Explanation
Problem Analysis We are given the equation x 2 + 5 x + x 2 + 5 x 36 = 0 and we want to find the values of x that satisfy this equation.
Substitution Let's make a substitution to simplify the equation. Let y = x 2 + 5 x . Then the equation becomes y + y 36 = 0.
Clearing the Fraction To solve for y , we can multiply both sides of the equation by y to get rid of the fraction: y 2 + 36 = 0.
Solving for y Now, we can solve for y :
y 2 = − 36 y = ± − 36 = ± 6 i .
Substituting Back Now we substitute y = x 2 + 5 x back into the solutions for y . We have two cases:
Case 1: x 2 + 5 x = 6 i Case 2: x 2 + 5 x = − 6 i
Solving Quadratic Equation - Case 1 Case 1: x 2 + 5 x = 6 i . This is a quadratic equation of the form x 2 + 5 x − 6 i = 0 . We can use the quadratic formula to solve for x :
x = 2 a − b ± b 2 − 4 a c where a = 1 , b = 5 , and c = − 6 i . Thus, x = 2 ( 1 ) − 5 ± 5 2 − 4 ( 1 ) ( − 6 i ) = 2 − 5 ± 25 + 24 i .
Solving Quadratic Equation - Case 2 Case 2: x 2 + 5 x = − 6 i . This is a quadratic equation of the form x 2 + 5 x + 6 i = 0 . We can use the quadratic formula to solve for x :
x = 2 a − b ± b 2 − 4 a c where a = 1 , b = 5 , and c = 6 i . Thus, x = 2 ( 1 ) − 5 ± 5 2 − 4 ( 1 ) ( 6 i ) = 2 − 5 ± 25 − 24 i .
Simplifying Square Roots and Finding Solutions We need to simplify the square roots 25 + 24 i and 25 − 24 i .
Let 25 + 24 i = a + bi , where a and b are real numbers. Squaring both sides, we get ( a + bi ) 2 = a 2 − b 2 + 2 abi = 25 + 24 i . Thus, a 2 − b 2 = 25 and 2 ab = 24 , so ab = 12 .
Similarly, let 25 − 24 i = c + d i , where c and d are real numbers. Squaring both sides, we get ( c + d i ) 2 = c 2 − d 2 + 2 c d i = 25 − 24 i . Thus, c 2 − d 2 = 25 and 2 c d = − 24 , so c d = − 12 .
Using a calculator, we find that 25 + 24 i ≈ 5.461 + 2.197 i and 25 − 24 i ≈ 5.461 − 2.197 i .
Therefore, the solutions are: x = 2 − 5 ± ( 5.461 + 2.197 i ) ≈ 0.231 + 1.099 i , − 5.231 − 1.099 i x = 2 − 5 ± ( 5.461 − 2.197 i ) ≈ 0.231 − 1.099 i , − 5.231 + 1.099 i
Final Answer The solutions to the equation x 2 + 5 x + x 2 + 5 x 36 = 0 are approximately 0.231 + 1.099 i , − 5.231 − 1.099 i , 0.231 − 1.099 i , and − 5.231 + 1.099 i .
Examples
Complex numbers and quadratic equations might seem abstract, but they're incredibly useful in electrical engineering. Imagine designing a circuit where you need to calculate impedance, which involves resistance, capacitance, and inductance. These quantities can be represented using complex numbers, and solving for the circuit's behavior often involves solving quadratic equations with complex coefficients. Understanding these concepts allows engineers to optimize circuit performance and ensure stability.
