Guided course 06:12Solving Quadratic Equations by the Square Root PropertyCallie1118views17rank1comments
Multiple ChoiceSolve the given quadratic equation using the square root property. (x−12)2−5=0\left(x-\frac12\right)^2-5=0(x−21)2−5=0278views1rank
Multiple ChoiceSolve the given quadratic equation using the square root property. 2x2−16=02x^2-16=02x2−16=0254views5rank
Textbook QuestionMatch the equation in Column I with its solution(s) in Column II. x^2 + 5 = 0234views
Textbook QuestionUse Choices A–D to answer each question. A. 3x^2 - 17x - 6 = 0 B. (2x + 5)^2 = 7 C. x^2 + x = 12 D. (3x - 1)(x - 7) = 0 Which equation is set up for direct use of the square root property? Solve it316views
Textbook QuestionAnswer each question. Answer each question. Answer each question. Unknown NumbersUse the following facts.If x represents an integer, then x+1 represents the next consecutive integer.If x represents an even integer, then x+2 represents the next consecutive even integer.If x represents an odd integer, then x+2 represents the next consecutive odd integer. Find two consecutive odd integers whose product is 63.217views
Textbook QuestionSolve each equation using the zero-factor property. See Example 1. x^2 + 2x - 8 = 0439views
Textbook QuestionAnswer each question. Answer each question. Answer each question. Unknown NumbersUse the following facts.If x represents an integer, then x+1 represents the next consecutive integer.If x represents an even integer, then x+2 represents the next consecutive even integer.If x represents an odd integer, then x+2 represents the next consecutive odd integer. The difference of the squares of two positive consecutive odd integers is 32. Find the integers.246views
Textbook QuestionSolve each equation using the zero-factor property. See Example 1. x^2 - 64 = 0230views
Textbook QuestionSolve each equation in Exercises 15–34 by the square root property. (x + 3)^2 = - 16204views
Textbook QuestionVolume of a Box. A rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in. long are cut from the four corners, and the flaps are folded upward to form an open box. If the volume of the box is 832 in.^3, what were the original dimensions of the piece of metal?277views
Textbook QuestionSolve each equation using the square root property. See Example 2. 27 - x^2 = 0254views
Textbook QuestionDimensions of a SquareWhat is the length of the side of a square if its area and perimeter are numerically equal?198views
Textbook QuestionSolve each equation using the square root property. See Example 2. (4x + 1)^2 = 20438views
Textbook QuestionSolve each equation using the square root property. See Example 2. (-2x + 5)^2 = -8214views
Textbook QuestionIn Exercises 35–46, determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then write and factor the trinomial. x^2 - 10x535views
Textbook QuestionSolve each equation using completing the square. See Examples 3 and 4. -2x^2 + 4x + 3 = 0266views
Textbook QuestionSolve each equation in Exercises 47–64 by completing the square. x^2 + 4x = 12251views
Textbook QuestionEvaluate the discriminant for each equation. Then use it to determine the number and type of solutions. -8x² + 10x = 7179views
Textbook QuestionSolve each equation using the quadratic formula. See Examples 5 and 6. x^2 - 3x - 2 = 0393views
Textbook QuestionSolve each equation using the quadratic formula. See Examples 5 and 6. 1/2x^2 + 1/4x - 3 = 0297views
Textbook QuestionSolve each equation in Exercises 47–64 by completing the square. 3x^2 - 2x - 2 = 0378views
Textbook QuestionSolve each equation using the quadratic formula. See Examples 5 and 6. (3x + 2)(x - 1) = 3x254views
Textbook QuestionSolve each equation in Exercises 66–67 by completing the square. 3x^2 -12x+11= 0295views
Textbook QuestionSolve each equation for the specified variable. (Assume no denominators are 0.) See Example 8. F = kMv^2/r , for v195views
Textbook QuestionIn Exercises 75–82, compute the discriminant. Then determine the number and type of solutions for the given equation. x^2 - 4x - 5 = 0505views
Textbook QuestionFor each equation, (a) solve for x in terms of y.. See Example 8. 4x^2 - 2xy + 3y^2 = 2205views
Textbook QuestionEvaluate the discriminant for each equation. Then use it to determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. (Do not solve the equation.) See Example 9. x^2 - 8x + 16 = 0352views
Textbook QuestionEvaluate the discriminant for each equation. Then use it to determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. (Do not solve the equation.) See Example 9. 9x^2 + 11x + 4 = 0392views
Textbook QuestionSolve each equation in Exercises 83–108 by the method of your choice. x^2 - 2x = 1235views
Textbook QuestionAnswer each question. Find the values of a, b, and c for which the quadratic equation. ax^2 + bx + c = 0 has the given numbers as solutions. (Hint: Use the zero-factor property in reverse.) i, -i361views
Textbook QuestionSolve each equation in Exercises 83–108 by the method of your choice. x^2 - 6x + 13 = 0296views
Textbook QuestionSolve each equation in Exercises 83–108 by the method of your choice. 3/(x - 3) + 5/(x - 4) = (x^2 - 20)/(x^2 - 7x + 12)212views
Textbook QuestionIn Exercises 115–122, find all values of x satisfying the given conditions. y1 = 2x/(x + 2), y2 = 3/(x + 4), and y1 + y2 = 1268views
Textbook QuestionIn Exercises 123–124, list all numbers that must be excluded from the domain of each rational expression. 3/(2x^2 + 4x - 9)297views
Textbook QuestionIn Exercises 127–130, solve each equation by the method of your choice. √2 x^2 + 3x - 2√2 = 0262views
Textbook QuestionThe rule for rewriting an absolute value equation without absolute value bars can be extended to equations with two sets of absolute value bars: If u and v represent algebraic expressions, then |u| = |v| is equivalent to u = v or u = - v. Use this to solve the equations in Exercises 77–84. |2x^2 - 4| = |2x^2|101views
Textbook QuestionUse the method described in Exercises 83–86, if applicable, and properties of absolute value to solve each equation or inequality. (Hint: Exercises 99 and 100 can be solved by inspection.) | x2 + 1 | - | 2x | = 024views