$\frac{2 \cdot \frac{5-r}{4}}{4}=\frac{1}{r}$
Answer (1)
After simplifying the given equation and attempting to solve for r , we arrive at a quadratic equation r 2 − 5 r + 8 = 0 . Applying the quadratic formula, we find that the discriminant is negative, indicating that there are no real solutions for r . Therefore, the equation has no real roots. no real solutions
Explanation
Analyze the problem We are given the equation 4 2 ⋅ 4 5 − r = r 1 Our goal is to solve for r .
Simplify the equation First, let's simplify the left-hand side of the equation: 4 2 ⋅ 4 5 − r = 4 2 5 − r = 2 ⋅ 4 5 − r = 8 5 − r So the equation becomes: 8 5 − r = r 1
Eliminate fractions Next, we multiply both sides of the equation by 8 r to eliminate the fractions: 8 r ⋅ 8 5 − r = 8 r ⋅ r 1 r ( 5 − r ) = 8
Expand the equation Expanding the left side, we get: 5 r − r 2 = 8
Rearrange to quadratic form Rearranging the terms to form a quadratic equation, we have: r 2 − 5 r + 8 = 0
Apply quadratic formula Now, we can use the quadratic formula to solve for r . The quadratic formula is given by: r = 2 a − b ± b 2 − 4 a c In our equation, a = 1 , b = − 5 , and c = 8 . Plugging these values into the quadratic formula, we get: r = 2 ( 1 ) − ( − 5 ) ± ( − 5 ) 2 − 4 ( 1 ) ( 8 ) r = 2 5 ± 25 − 32 r = 2 5 ± − 7 Since the discriminant (the value inside the square root) is negative, there are no real solutions for r .
Double Check Since there are no real solutions, we made a mistake somewhere. Let's go back to step 2: 4 2 ⋅ 4 5 − r = r 1 8 5 − r = r 1 r ( 5 − r ) = 8 5 r − r 2 = 8 r 2 − 5 r + 8 = 0 The quadratic formula is: r = 2 a − b ± b 2 − 4 a c r = 2 ( 1 ) 5 ± ( − 5 ) 2 − 4 ( 1 ) ( 8 ) r = 2 5 ± 25 − 32 r = 2 5 ± − 7 There are no real roots.
Final Check Let's re-examine the original equation. 4 2 ⋅ 4 5 − r = r 1 16 2 ( 5 − r ) = r 1 8 5 − r = r 1 r ( 5 − r ) = 8 5 r − r 2 = 8 r 2 − 5 r + 8 = 0 Using the quadratic formula: r = 2 5 ± 25 − 4 ( 1 ) ( 8 ) r = 2 5 ± 25 − 32 r = 2 5 ± − 7 Since the discriminant is negative, there are no real solutions.
Conclusion Since the discriminant is negative, there are no real solutions to the equation. Therefore, there is no real value of r that satisfies the given equation.
Examples
When designing electrical circuits, engineers often encounter equations that need to be solved to determine the values of components like resistors or capacitors. The given equation is similar to those that might arise when analyzing the behavior of a circuit with feedback or resonance. Although this particular equation has no real solutions, understanding how to manipulate and solve such equations is crucial for ensuring the stability and performance of electronic devices.
