Rewrite the equation in standard form: z 2 − 16 z + 104 = 0 .
Apply the quadratic formula: z = 2 a − b ± b 2 − 4 a c .
Substitute a = 1 , b = − 16 , and c = 104 into the formula.
Simplify to find the complex solutions: z = 8 ± 2 i 10 .
Explanation
Problem Analysis We are given the quadratic equation z 2 − 16 z + 30 = − 74 . Our goal is to solve for z , simplifying all answers, including any non-real solutions.
Rewriting the Equation First, we rewrite the equation in the standard quadratic form a z 2 + b z + c = 0 . To do this, we add 74 to both sides of the equation:
z 2 − 16 z + 30 + 74 = − 74 + 74
z 2 − 16 z + 104 = 0
Applying the Quadratic Formula Now we can use the quadratic formula to solve for z . The quadratic formula is given by:
z = 2 a − b ± b 2 − 4 a c
In our equation, a = 1 , b = − 16 , and c = 104 . Plugging these values into the quadratic formula, we get:
z = 2 ( 1 ) − ( − 16 ) ± ( − 16 ) 2 − 4 ( 1 ) ( 104 )
z = 2 16 ± 256 − 416
z = \frac{16 \pm \sqrt{-160}}{2}$ 4. Simplifying the Solutions Since the discriminant (the value inside the square root) is negative, we will have complex solutions. We can simplify the square root of -160 as follows: \sqrt{-160} = \sqrt{16 \times -10} = \sqrt{16} \times \sqrt{-10} = 4\sqrt{-10} = 4i\sqrt{10} S o , o u rso l u t i o n s b eco m e : z = \frac{16 \pm 4i\sqrt{10}}{2} z = 8 \pm 2i\sqrt{10}
Final Answer Therefore, the two solutions for z are 8 + 2 i 10 and 8 − 2 i 10 .
Stating the Solutions The solutions are z = 8 + 2 i 10 and z = 8 − 2 i 10 .
Examples
Quadratic equations are used in various fields such as physics, engineering, and economics. For example, in physics, projectile motion can be modeled using quadratic equations to determine the trajectory of an object. In engineering, quadratic equations can be used to design bridges and buildings. In economics, they can be used to model cost and revenue functions to optimize profits. Understanding how to solve quadratic equations is essential for solving real-world problems in these fields.
To solve the equation z 2 − 16 z + 30 = − 74 , we rewrite it in standard form and use the quadratic formula. The solutions are z = 8 + 2 i 10 and z = 8 − 2 i 10 . This indicates that the equation has two complex solutions.
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