$x^2+\frac{5 x+36}{x^2+5 x}=0$
Answer (2)
To solve the equation x 2 + x 2 + 5 x 5 x + 36 = 0 , we eliminate the fraction by multiplying through by x 2 + 5 x and derive the quartic equation x 4 + 5 x 3 + 5 x + 36 = 0 . The approximate real roots are x ≈ − 4.90 and x ≈ − 2.06 .
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Multiply both sides by x 2 + 5 x to get rid of the fraction: x 4 + 5 x 3 + 5 x + 36 = 0 .
Use a computational tool to find the roots of the quartic equation.
Identify the real roots: approximately − 4.9025 and − 2.0599 .
Round the real roots to two decimal places: x ≈ − 4.90 , − 2.06 .
Explanation
Initial Equation We are given the equation x 2 + x 2 + 5 x 5 x + 36 = 0
Eliminating the Fraction First, we multiply both sides of the equation by x 2 + 5 x to eliminate the fraction. Note that x cannot be 0 or -5, since the denominator x 2 + 5 x = x ( x + 5 ) would be zero. Thus, x 2 ( x 2 + 5 x ) + 5 x + 36 = 0
Expanding the Equation Expanding the equation, we obtain x 4 + 5 x 3 + 5 x + 36 = 0
Finding the Roots Let P ( x ) = x 4 + 5 x 3 + 5 x + 36 . We look for rational roots using the Rational Root Theorem. Possible rational roots are ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 9 , ± 12 , ± 18 , ± 36 . However, testing these roots can be time-consuming, and it's not guaranteed that we'll find any rational roots. Instead, we can use a computational tool to find the roots of the polynomial directly. The roots are approximately: − 4.9025 , − 2.0599 , 0.9812 + 1.6130 i , 0.9812 − 1.6130 i
Verifying the Roots The real roots are approximately x = − 4.9025 and x = − 2.0599 . We should check if these roots satisfy the original equation. For x = − 4.9025 :
( − 4.9025 ) 2 + ( − 4.9025 ) 2 + 5 ( − 4.9025 ) 5 ( − 4.9025 ) + 36 ≈ 24.0345 + 24.0345 − 24.5125 11.4875 ≈ 24.0345 + − 0.478 11.4875 ≈ 24.0345 − 24.0324 ≈ 0.0021 ≈ 0 For x = − 2.0599 :
( − 2.0599 ) 2 + ( − 2.0599 ) 2 + 5 ( − 2.0599 ) 5 ( − 2.0599 ) + 36 ≈ 4.2431 + 4.2431 − 10.2995 25.7005 ≈ 4.2431 + − 6.0564 25.7005 ≈ 4.2431 − 4.2431 ≈ 0
Final Answer Therefore, the real roots of the equation are approximately x = − 4.9025 and x = − 2.0599 .
We can round these to two decimal places: x ≈ − 4.90 and x ≈ − 2.06 .
Examples
Understanding polynomial roots is crucial in many engineering applications, such as control systems design. For instance, when designing a robotic arm, engineers use polynomial equations to model the arm's movement and stability. The roots of these polynomials help determine the arm's stable operating points and prevent unwanted oscillations. Similarly, in signal processing, roots of polynomials are used to analyze and design filters that remove noise from audio or image signals. By finding the roots, engineers can optimize the filter's performance to achieve the desired signal clarity.
