Topics
>
Algebra 2>
Quadratic Equations and Functions>
Completing the Square>
Completing the Square 2Completing the Square 2
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 5 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- 5
- Answered
- Review
-
Question 1 of 5
1. Question
Solve by completing the square`2x^2-12x=-8`Hint
Help VideoCorrect
Correct!
Incorrect
Completing the square is done by taking the coefficient of `x`, halving it and then squaring it. Then we add the new value to both sides of the equation.
Simplify the equation by dividing both sides by `2``2x^2-12x` `=` `-8` `(2x^2-12x)``-:2` `=` `-8` `-:2` `x^2-6x` `=` `-4` Take the coefficient of the middle term, divide it by two and then square it.`x^2``-6``x` `=` `-4` Coefficient of the middle term `-6``-:2` `=` `-3` Divide it by `2` `(-3)^2` `=` `9` Square This number will make a perfect square on the left side.Add `9` to both sides of the equation to keep the balance, then form a square of a binomial`x^2-6x` `=` `-4` `x^2-6x` `+9` `=` `-4` `+9` `(x-3)^2` `=` `5` Finally, take the square root of both sides and continue solving for `x`.`(x-3)^2` `=` `5` `x-3` `=` `sqrt(5)` `x-3` `+3` `=` `+-sqrt(5)` `+3` Add `3` to both sides `x` `=` `3+-sqrt5` The roots can also be written individually`=` `3+sqrt5` `=` `3-sqrt5` `3+-sqrt5` -
Question 2 of 5
2. Question
Solve by completing the square`4x^2+8x-7=0`Hint
Help VideoCorrect
Great Work!
Incorrect
Completing the square is done by taking the coefficient of `x`, halving it and then squaring it. Then we add the new value to both sides of the equation.Divide the equation by `4` to reduce the coefficient of `x^2``4x^2+8x-7` `=` `0` `4x^2+8x-7``-:4` `=` `0``-:4` `x^2+2x-7/4` `=` `0` Take the coefficient of the middle term, divide it by two and then square it.`x^2+``2``x-7/4` `=` `0` Coefficient of the middle term `2``-:2` `=` `1` Divide it by `2` `(1)^2` `=` `1` Square This number will make a perfect square on the left side.Add and subtract `1` to the left side of the equation to keep the balance, then form a square of a binomial`x^2+2x-7/4` `=` `0` `x^2+2x` `+1-1``-7/4` `=` `0` `(x+1)^2-1-7/4` `=` `0` `(x+1)^2-11/4` `=` `0` Move the constant to the right`(x+1)^2-11/4` `=` `0` `(x+1)^2-11/4` `+11/4` `=` `0` `+11/4` Add `11/4` to both sides `(x+1)^2` `=` `11/4` Finally, take the square root of both sides and continue solving for `x`.`(x+1)^2` `=` `11/4` `sqrt((x+1)^2)` `=` `sqrt(11/4)` `x+1` `=` `+-(sqrt11)/2` `x+1` `-1` `=` `+-(sqrt11)/2` `-1` Subtract `1` from both sides `x` `=` `-1+-(sqrt11)/2` The roots can also be written individually`=` `-1+(sqrt11)/2` `=` `-1-(sqrt11)/2` `-1+-(sqrt11)/2` -
Question 3 of 5
3. Question
Solve by completing the square`-4x^2+21x=x+5`Hint
Help VideoCorrect
Nice Job!
Incorrect
Completing the square is done by taking the coefficient of `x`, halving it and then squaring it. Then we add the new value to both sides of the equation.Simplify the equation and divide the equation by `-4` to reduce the coefficient of `x^2``-4x^2+21x` `=` `x+5` `-4x^2+21x` `-x` `=` `x+5` `-x` Subtract `x` from both sides `-4x^2+20x` `=` `5` `(-4x^2+20x)``-:(-4)` `=` `5``-:(-4)` Divide both sides by `-4` `x^2-5x` `=` `-5/4` Take the coefficient of the middle term, divide it by two and then square it.`x^2``-5``x` `=` `-5/4` Coefficient of the middle term `-5``-:2` `=` `-5/2` Divide it by `2` `(-5/2)^2` `=` `25/4` Square This number will make a perfect square on the left side.Add `25/4` to both sides to keep the balance, then form a square of a binomial`x^2-5x` `=` `-5/4` `x^2-5x` `+25/4` `=` `-5/4` `+25/4` `(x-5/2)^2` `=` `20/4` `(x-5/2)^2` `=` `5` Finally, take the square root of both sides and continue solving for `x`.`(x-5/2)^2` `=` `5` `sqrt((x-5/2)^2)` `=` `sqrt5` `x-5/2` `=` `+-sqrt5` `x-5/2` `+5/2` `=` `+-sqrt5` `+5/2` Add 5/2 to both sides `x` `=` `5/2+-sqrt5` The roots can also be written individually`=` `5/2+sqrt5` `=` `5/2-sqrt5` `5/2+-sqrt5` -
Question 4 of 5
4. Question
Solve by completing the square`9x^2+10x-100=4x^2+15`Hint
Help VideoCorrect
Fantastic!
Incorrect
Completing the square is done by taking the coefficient of `x`, halving it and then squaring it. Then we add the new value to both sides of the equation.Simplify the equation, making sure that `x^2` has `1` as a coefficient`9x^2+10x-100` `=` `4x^2+15` `9x^2+10x-100` `-4x^2` `=` `4x^2+15` `-4x^2` Subtract `4x^2` from both sides `5x^2+10x-100` `=` `15` `5x^2+10x-100` `+100` `=` `15` `+100` Add `100` to both sides `5x^2+10x` `=` `115` `5x^2+10x``-:5` `=` `115``-:5` Divide both sides by `5` `x^2+2x` `=` `23` Take the coefficient of the middle term, divide it by two and then square it.`x^2+``2``x` `=` `23` Coefficient of the middle term `2``-:2` `=` `1` Divide it by `2` `(1)^2` `=` `1` Square This number will make a perfect square on the left side.Add `1` to both sides to keep the balance, then form a square of a binomial`x^2+2x` `=` `23` `x^2+2x` `+1` `=` `23` `+1` `(x+1)^2` `=` `24` Finally, take the square root of both sides and continue solving for `x`.`(x+1)^2` `=` `24` `sqrt((x+1)^2)` `=` `sqrt(24)` `x+1` `=` `+-2sqrt6` `x+1` `-1` `=` `+-2sqrt6` `-1` Subtract `1` from both sides `x` `=` `-1+-2sqrt6` The roots can also be written individually`=` `-1+2sqrt6` `=` `-1-2sqrt6` `-1+-2sqrt6` -
Question 5 of 5
5. Question
Find the vertex of the function:`y=3x^2-9x+4`Hint
Help VideoCorrect
Correct!
Incorrect
Vertex Form
`y=a(x-h)^2+k`where `(h,k)` is the vertexCompleting the square is done by taking the coefficient of `x`, halving it and then squaring it. Then we add the new value to both sides of the equation.To find the vertex, transform the given function into vertex formStart by leaving `x` terms on the right side and factoring it out`y` `=` `3x^2-9x+4` `y``-4` `=` `3x^2-9x+4``-4` Subtract `4` from both sides `y-4` `=` `3x^2-9x` `y-4` `=` `3(x^2-3x)` Factor out `3` Take the coefficient of the `x` term, divide it by two and then square it.`y-4` `=` `3(x^2``-3``x)` Coefficient of the `x` term `=` $$\frac{\color{#00880A}{-3}}{2}$$ Divide it by `2` $$({\frac{\color{#00880A}{-3}}{2}})^2$$ `=` `9/4` Square This number will make the `x` terms a perfect square.Add and subtract `9/4` to the grouping of `x` terms to keep the balance.`y-4` `=` `3(x^2-3x)` `y-4` `=` `3(x^2-3x+``9/4``-``9/4``)` Add and subtract `9/4` Now, transform the grouping of `x` terms into a square of a binomial.[show cross method with two `x`’s on the left and two `-3/2`s on the right]`y-4` `=` `3[(x-3/2)(x-3/2)-9/4]` `y-4` `=` `3[(x-3/2)^2-9/4]` Now, distribute `3` and leave `y` on the left side`y-4` `=` `3[(x-3/2)^2-9/4]` `y-4` `=` `3(x-3/2)^2-3(9/4)` Distribute `3` `y-4` `=` `3(x-3/2)^2-27/4` `y-4``+4` `=` `3(x-3/2)^2-27/4``+4` Add `4` to both sides `y` `=` `3(x-3/2)^2-11/4` Finally, the function is in vertex formCompare the function to the general vertex form to get the vertex`y` `=` `a(x-h)^2+k` `y` `=` `3(x-3/2)^2-11/4` `h` `=` `3/2` `=` `1 1/2` `k` `=` `-11/4` `=` `-2 3/4` `(1 1/2, -2 3/4)`
Quizzes
- Solve Quadratics by Factoring
- The Quadratic Formula
- Completing the Square 1
- Completing the Square 2
- Intro to Quadratic Functions (Parabolas) 1
- Intro to Quadratic Functions (Parabolas) 2
- Intro to Quadratic Functions (Parabolas) 3
- Graph Quadratic Functions in Standard Form 1
- Graph Quadratic Functions in Standard Form 2
- Graph Quadratic Functions by Completing the Square
- Graph Quadratic Functions in Vertex Form
- Write a Quadratic Equation from the Graph
- Write a Quadratic Equation Given the Vertex and Another Point
- Quadratic Inequalities 1
- Quadratic Inequalities 2
- Quadratics Word Problems 1
- Quadratics Word Problems 2
- Quadratic Identities
- Graphing Quadratics Using the Discriminant
- Positive and Negative Definite
- Applications of the Discriminant 1
- Applications of the Discriminant 2
- Combining Methods for Solving Quadratic Equations