d) (6k 5)2 20 0 6k 5 2i15 Simplify 120. k The check is left to the student. The solution set is e 5 2i15 YOU TRY 3 Solve. (6k 5)2 20 Subtract 20 from each side. 6k 5 120 Square root property 6k 5 2i15 Add 5 to each side. 5 2i15 6 Divide by 6. a) (7q 1)2 36 b) (5a 3)2 24 c) (c 7)2 100 0 d) (2y 3)2 5 = 23 6 , 5 2i15 6 f . Did you notice in Examples 1c), 3c), and 3d) that a complex number and its conjugate were the solutions to the equations? This will always be true provided that the variables in the equation have real number coeffi cients. Note If a bi is a solution of a quadratic equation having only real coefficients, then a bi is also a solution. 3 Use the Distance Formula In mathematics, we sometimes need to fi nd the distance between two points in a plane. The distance formula enables us to do that. We can use the Pythagorean theorem and the square root property to develop the distance formula. Suppose we want to fi nd the distance between any two points with coordinates (x1, y1) and (x2, y2) as pictured here. We also include the point (x2, y1) in our drawing so that we get a right triangle. The lengths of the legs are a and b. The length of the hypotenuse is c. Our goal is to fi nd the distance between (x1, y1) and (x2, y2), which is the same as fi nding the length of c. How long is side a? |x2 x1| How long is side b? |y2 y1| c b x2 y2 x1 y1 x2 y1 a x y The Pythagorean theorem states that a2 b2 c2. Substitute x2 x1 for a and y2 y1 for b, then solve for c. a2 b2 c2 Pythagorean theorem x2 x12 y2 y12 c2 Substitute values. 2(x2 x1)2 ( y2 y1)2 c Solve for c using the square root property. The distance between the points (x1, y1) and (x2, y2) is c2(x2 x1)2 ( y2 y1)2. We want only the positive square root since c is a length. Distances will always be a positive number or zero. www.mhhe.com/messersmith SECTION 10.1 The Square Root Property and Completing the Square 615
messersmith_power_intermediate_algebra_1e_ch4_7_10
To see the actual publication please follow the link above