Prerequisite Skills This lesson requires the use of the following skills: simplifying radicals working with complex numbers Introduction You can determine how far a ladder will extend from the base of a wall by creating a quadratic equation and then taking the square root. To find this length, you only need to find the positive square root because a negative distance would not make sense in this situation. For certain types of quadratics, we can solve by taking the square root, but in most problems, we need to take both the positive and negative square root. Key Concepts The imaginary unit i represents the non-real value i = 1. i is the number whose square is 1. We define i so that i = 1 and i 2 = 1. An imaginary number is any number of the form bi, where b is a real number, i = 1, and b 0. A complex number is a number with a real component and an imaginary component. Complex numbers can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit. For example, 5 + 3i is a complex number. 5 is the real component and 3i is the imaginary component. Real numbers are the set of all rational and irrational numbers. Real numbers do not contain an imaginary component. Real numbers are rational numbers when they can be written as m, where both m and n are n integers and n 0. Rational numbers can also be written as a decimal that ends or repeats. The real number 0.4 is a rational number because it can be written as the fraction 2 5. Real numbers are irrational when they cannot be written as m, where m and n are integers n and n 0. Irrational numbers cannot be written as a decimal that ends or repeats. The real number 3 is an irrational number because it cannot be written as the ratio of two integers. Other examples of irrational numbers include 2 and π. U5-39
A quadratic equation is an equation that can be written in the form ax 2 + bx + c = 0, where a 0. Quadratic equations can have no real solutions, one real solution, or two real solutions. When a quadratic has no real solutions, it has two complex solutions. Quadratic equations that contain only a squared term and a constant can be solved by taking the square root of both sides. These equations can be written in the form x 2 = c, where c is a constant. When we take the square root of both sides, we need to remember that a number and its opposite have the same square. Therefore, rather than simply taking the positive square root, we need to take the positive and negative square root. For x 2 = c, we find that x=± c. We can use a similar method to solve quadratic equations in the form (ax + b) 2 = c. c tells us the number and type of solutions for the equation. c Number and type of solutions Negative Two complex solutions 0 One real, rational solution Positive and a perfect square Two real, rational solutions Positive and not a perfect square Two real, irrational solutions Common Errors/Misconceptions forgetting to use ± and therefore forgetting that there may be two solutions taking the square root before isolating the squared term U5-40
Guided Practice 5.2.1 Example 1 Solve 2x 2 5 = 195 for x. 1. Isolate x 2. 2x 2 5 = 195 Original equation 2x 2 = 200 Add 5 to both sides. x 2 = 100 Divide both sides by 2. 2. Use a square root to find all possible solutions to the equation. Take the square root of both sides. Remember that both 10 2 and ( 10) 2 equal 100. x =± 100 =± 10 The equation 2x 2 5 = 195 has two solutions, 10 and 10. U5-41
Example 2 Solve (x 1) 2 + 15 = 1 for x. 1. Isolate the squared binomial. (x 1) 2 + 15 = 1 Original equation (x 1) 2 = 16 Subtract 15 from both sides. 2. Use a square root to isolate the binomial. Take the square root of both sides. Remember to use the ± sign. x 1=± 16 3. Simplify the square root. There is a negative number under the radical, so the answer will be a complex number. x 1=± 16 Equation x 1=± ( 1)( 16) x 1=± 1 16 x 1 = ±4i Write 16 as a product of a perfect square and 1. Product Property of Square Roots Simplify. 4. Isolate x. x 1 = ±4i x = 1 ± 4i Equation Add 1 to both sides. The equation (x 1) 2 + 15 = 1 has two solutions, 1 ± 4i. U5-42
Example 3 Solve 4(x + 3) 2 10 = 6 for x. 1. Isolate the squared binomial. 4(x + 3) 2 10 = 6 Original equation 4(x + 3) 2 = 4 Add 10 to both sides. (x + 3) 2 = 1 Divide both sides by 4. x + 3=± 1 Take the square root of both sides. x + 3 = ±1 2. Isolate x. x + 3 = ±1 x = 3 ± 1 Equation Subtract 3 from both sides. 3. Split the answer into two separate expressions and evaluate. x = 3 + 1 = 2 x = 3 1 = 4 The equation 4(x + 3) 2 10 = 6 has two solutions, 2 and 4. U5-43