The solutions for the first equation k 2 + 10 = 0 are k = ± i 10 with the factors being ( k − i 10 ) ( k + i 10 ) = 0 . For the second equation 4 x 2 + 29 = 0 , the solutions are x = ± i 2 29 with the factors ( 2 x − i 29 ) ( 2 x + i 29 ) = 0 .
;
Solve k 2 + 10 = 0 to find k = ± i 10 .
Factor the equation as ( k − i 10 ) ( k + i 10 ) = 0 .
Solve 4 x 2 + 29 = 0 to find x = ± i 2 29 .
Factor the equation as ( 2 x − i 29 ) ( 2 x + i 29 ) = 0 .
The solutions are k = ± i 10 and x = ± i 2 29 .
Explanation
Understanding the Problem We are given two equations: k 2 + 10 = 0 and 4 x 2 + 29 = 0 . We need to solve these equations by factoring and find their solutions.
Solving the First Equation Let's solve the first equation, k 2 + 10 = 0 . We can rewrite this as k 2 = − 10 . Taking the square root of both sides, we get k = ± − 10 . Since we have a negative number under the square root, we can express the solution in terms of the imaginary unit i , where i = − 1 . Thus, k = ± i 10 . The factors are ( k − i 10 ) and ( k + i 10 ) .
Solving the Second Equation Now let's solve the second equation, 4 x 2 + 29 = 0 . We can rewrite this as 4 x 2 = − 29 . Dividing both sides by 4, we get x 2 = − 4 29 . Taking the square root of both sides, we get x = ± − 4 29 . Again, we have a negative number under the square root, so we can express the solution in terms of the imaginary unit i . Thus, x = ± i 2 29 . The factors are ( 2 x − i 29 ) and ( 2 x + i 29 ) .
Final Solutions Therefore, the solutions for the first equation are k = i 10 and k = − i 10 , and the solutions for the second equation are x = i 2 29 and x = − i 2 29 .
Examples
Complex numbers might seem abstract, but they're incredibly useful in electrical engineering. For example, when analyzing AC circuits, impedance (a measure of opposition to current) is often expressed as a complex number. Solving equations with complex numbers helps engineers design and understand these circuits, ensuring our electronic devices work correctly.