Guided course 06:12Solving Quadratic Equations by the Square Root PropertyCallie1129views17rank1comments
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=0283views1rank
Multiple ChoiceSolve the given quadratic equation using the square root property. 2x2−16=02x^2-16=02x2−16=0257views5rank
Textbook QuestionMatch the equation in Column I with its solution(s) in Column II. x^2 + 5 = 0236views
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 it319views
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 63.221views
Textbook QuestionSolve each equation using the zero-factor property. See Example 1. x^2 + 2x - 8 = 0444views
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.248views
Textbook QuestionSolve each equation using the zero-factor property. See Example 1. x^2 - 64 = 0235views
Textbook QuestionSolve each equation in Exercises 15–34 by the square root property. (x + 3)^2 = - 16206views
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?280views
Textbook QuestionSolve each equation using the square root property. See Example 2. 27 - x^2 = 0259views
Textbook QuestionDimensions of a SquareWhat is the length of the side of a square if its area and perimeter are numerically equal?201views
Textbook QuestionSolve each equation using the square root property. See Example 2. (4x + 1)^2 = 20441views
Textbook QuestionSolve each equation using the square root property. See Example 2. (-2x + 5)^2 = -8219views
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 - 10x541views
Textbook QuestionSolve each equation using completing the square. See Examples 3 and 4. -2x^2 + 4x + 3 = 0269views
Textbook QuestionSolve each equation in Exercises 47–64 by completing the square. x^2 + 4x = 12252views
Textbook QuestionEvaluate the discriminant for each equation. Then use it to determine the number and type of solutions. -8x² + 10x = 7183views
Textbook QuestionSolve each equation using the quadratic formula. See Examples 5 and 6. x^2 - 3x - 2 = 0403views
Textbook QuestionSolve each equation using the quadratic formula. See Examples 5 and 6. 1/2x^2 + 1/4x - 3 = 0300views
Textbook QuestionSolve each equation in Exercises 47–64 by completing the square. 3x^2 - 2x - 2 = 0387views
Textbook QuestionSolve each equation using the quadratic formula. See Examples 5 and 6. (3x + 2)(x - 1) = 3x258views
Textbook QuestionSolve each equation in Exercises 66–67 by completing the square. 3x^2 -12x+11= 0297views
Textbook QuestionSolve each equation for the specified variable. (Assume no denominators are 0.) See Example 8. F = kMv^2/r , for v199views
Textbook QuestionIn Exercises 75–82, compute the discriminant. Then determine the number and type of solutions for the given equation. x^2 - 4x - 5 = 0511views
Textbook QuestionFor each equation, (a) solve for x in terms of y.. See Example 8. 4x^2 - 2xy + 3y^2 = 2210views
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 = 0355views
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 = 0398views
Textbook QuestionSolve each equation in Exercises 83–108 by the method of your choice. x^2 - 2x = 1238views
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, -i366views
Textbook QuestionSolve each equation in Exercises 83–108 by the method of your choice. x^2 - 6x + 13 = 0298views
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)214views
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 = 1271views
Textbook QuestionIn Exercises 123–124, list all numbers that must be excluded from the domain of each rational expression. 3/(2x^2 + 4x - 9)300views
Textbook QuestionIn Exercises 127–130, solve each equation by the method of your choice. √2 x^2 + 3x - 2√2 = 0268views
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|104views
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 | = 026views