Completing the Square 1
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Question 1 of 5
1. Question
Solve by completing the square`x^2-12x=-22`Hint
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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.
Take the coefficient of the middle term, divide it by two and then square it.`x^2``-12``x` `=` `-22` Coefficient of the middle term `-12``-:2` `=` `-6` Divide it by `2` `(-6)^2` `=` `36` Square This number will make the left side a perfect square.Add `36` to both sides of the equation to keep the balance.`x^2-12x` `=` `-22` `x^2-12x` `+36` `=` `-22` `+36` Add `36` to both sides `x^2-12x+36` `=` `14` Now, transform the left side into a square of a binomial by factoring or using cross method.
`(x-6)(x-6)` `=` `14` `(x-6)^2` `=` `14` Finally, take the square root of both sides and continue solving for `x`.`(x-6)^2` `=` `14` `sqrt((x-6)^2)` `=` `sqrt14` Take the square root `x-6` `=` `+-sqrt(14)` Square rooting a number gives a plus and minus solution `x-6` `+6` `=` `+-sqrt(14)` `+6` Add `6` to both sides `x` `=` `6+-sqrt(14)` Simplify The roots can also be written individually`=` $$6+\sqrt14$$ `=` $$6-\sqrt14$$ `x=6+-sqrt(14)` -
Question 2 of 5
2. Question
Solve by completing the square`x^2+6x+13=0`Hint
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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.Take the coefficient of the middle term, divide it by two and then square it.`x^2+``6``x+13` `=` `0` 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 and subtract `9` to the left side of the equation to keep the balance, then form a square of a binomial`x^2+6x+13` `=` `0` `x^2+6x` `+9-9``+13` `=` `0` `(x+3)^2-9+13` `=` `0` `(x+3)^2+4` `=` `0` Move the constant to the right`(x+3)^2+4` `=` `0` `(x+3)^2+4` `-4` `=` `0` `-4` Subtract `4` from both sides `(x+3)^2` `=` `-4` Finally, take the square root of both sides and continue solving for `x`.`(x+3)^2` `=` `-4` `sqrt((x+3)^2)` `=` `sqrt(-4)` Remember that a negative value inside a surd will give out imaginary roots. Therefore, this equation has no real rootsNo real roots -
Question 3 of 5
3. Question
Solve by completing the square`x^2+3x-6=0`Hint
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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.Take the coefficient of the middle term, divide it by two and then square it.`x^2+``3``x-6` `=` `0` Coefficient of the middle term `3``-:2` `=` `3/2` Divide it by `2` `(3/2)^2` `=` `9/4` Square This number will make a perfect square on the left side.Add and subtract `9/4` to the left side of the equation to keep the balance, then form a square of a binomial`x^2+3x-6` `=` `0` `x^2+3x` `+9/4-9/4``-6` `=` `0` `(x+3/2)^2-9/4-6` `=` `0` `(x+3/2)^2-33/4` `=` `0` Move the constant to the right`(x+3/2)^2-33/4` `=` `0` `(x+3/2)^2-33/4` `+33/4` `=` `0` `+33/4` Add `33/4` to both sides `(x+3/2)^2` `=` `33/4` Finally, take the square root of both sides and continue solving for `x`.`(x+3/2)^2` `=` `33/4` `sqrt((x+3/2)^2)` `=` `sqrt(33/4)` `x+3/2` `=` `(+-sqrt33)/2` `x+3/2` `-3/2` `=` `(+-sqrt33)/2` `-3/2` Subtract `3/2` from both sides `x` `=` `(-3+-sqrt33)/2` The roots can also be written individually`=` `(-3+sqrt33)/2` `=` `(-3-sqrt33)/2` `(-3+-sqrt33)/2` -
Question 4 of 5
4. Question
Solve by completing the square`k^2-4k=2k+18`Hint
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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.Move `k` terms to the left`k^2-4k` `=` `2k+18` `k^2-4k` `-2k` `=` `2k+18` `-2k` Subtract `2k` from both sides `k^2-6k` `=` `18` Take the coefficient of the middle term, divide it by two and then square it.`k^2``-6``k` `=` `18` 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`k^2-6k` `=` `18` `k^2-6k` `+9` `=` `18` `+9` `(x-3)^2` `=` `27` Finally, take the square root of both sides and continue solving for `x`.`(x-3)^2` `=` `27` `sqrt((x-3)^2)` `=` `sqrt(27)` `x-3` `=` `+-3sqrt3` `x-3` `+3` `=` `+-3sqrt3` `+3` Add `3` to both sides `x` `=` `3+-3sqrt3` The roots can also be written individually`=` `3+3sqrt3` `=` `3-3sqrt3` `3+-3sqrt3` -
Question 5 of 5
5. Question
Solve by completing the square`u^2+1.8u-2.2=0`Hint
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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.Take the coefficient of the middle term, divide it by two and then square it.`u^2+``1.8``u-2.2` `=` `0` Coefficient of the middle term `1.8``-:2` `=` `0.9` Divide it by `2` `(0.9)^2` `=` `0.81` Square This number will make a perfect square on the left side.Add and subtract `0.81` to the left side of the equation to keep the balance, then form a square of a binomial`u^2+1.8u-2.2` `=` `0` `u^2+1.8u` `+0.81-0.81``-2.2` `=` `0` `(u+0.9)^2-0.81-2.2` `=` `0` `(u+0.9)^2-3.01` `=` `0` Move the constant to the right`(u+0.9)^2-3.01` `=` `0` `(u+0.9)^2-3.01` `+3.01` `=` `0` `+3.01` Add `3.01` to both sides `(u+0.9)^2` `=` `3.01` Finally, take the square root of both sides and continue solving for `x`.`(u+0.9)^2` `=` `3.01` `sqrt((u+0.9)^2)` `=` `sqrt(3.01)` `u+0.9` `=` `+-sqrt(3.01)` `u+0.9` `-0.9` `=` `+-sqrt(3.01)` `-0.9` Subtract `0.9` from both sides `u` `=` `-0.9+-sqrt(3.01)` The roots can also be written individually`=` `-0.9+sqrt(3.01)` `=` `-0.9-sqrt(3.01)` `-0.9+-sqrt(3.01)`
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- Sum & Product of Roots 2
- Sum & Product of Roots 3
- Sum & Product of Roots 4
- Solving Equations by Factoring 1
- Solving Equations Using the Quadratic Formula
- Completing the Square 1
- Completing the Square 2
- Intro to Quadratic Functions (Parabolas) 1
- Intro to Quadratic Functions (Parabolas) 2
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- 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
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- 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
- Solving Reducible Equations