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Completing the Square>
Completing the Square 1Completing the Square 1
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Question 1 of 5
1. Question
Solve by completing the square`x^212x=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^212x` `=` `22` `x^212x` `+36` `=` `22` `+36` Add `36` to both sides `x^212x+36` `=` `14` Now, transform the left side into a square of a binomial by factoring or using cross method.`(x6)(x6)` `=` `14` `(x6)^2` `=` `14` Finally, take the square root of both sides and continue solving for `x`.`(x6)^2` `=` `14` `sqrt((x6)^2)` `=` `sqrt14` Take the square root `x6` `=` `+sqrt(14)` Square rooting a number gives a plus and minus solution `x6` `+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` `+99``+13` `=` `0` `(x+3)^29+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+3x6=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``x6` `=` `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+3x6` `=` `0` `x^2+3x` `+9/49/4``6` `=` `0` `(x+3/2)^29/46` `=` `0` `(x+3/2)^233/4` `=` `0` Move the constant to the right`(x+3/2)^233/4` `=` `0` `(x+3/2)^233/4` `+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` `=` `(3sqrt33)/2` `(3+sqrt33)/2` 
Question 4 of 5
4. Question
Solve by completing the square`k^24k=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^24k` `=` `2k+18` `k^24k` `2k` `=` `2k+18` `2k` Subtract `2k` from both sides `k^26k` `=` `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^26k` `=` `18` `k^26k` `+9` `=` `18` `+9` `(x3)^2` `=` `27` Finally, take the square root of both sides and continue solving for `x`.`(x3)^2` `=` `27` `sqrt((x3)^2)` `=` `sqrt(27)` `x3` `=` `+3sqrt3` `x3` `+3` `=` `+3sqrt3` `+3` Add `3` to both sides `x` `=` `3+3sqrt3` The roots can also be written individually`=` `3+3sqrt3` `=` `33sqrt3` `3+3sqrt3` 
Question 5 of 5
5. Question
Solve by completing the square`u^2+1.8u2.2=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``u2.2` `=` `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+1.8u2.2` `=` `0` `u^2+1.8u` `+0.810.81``2.2` `=` `0` `(u+0.9)^20.812.2` `=` `0` `(u+0.9)^23.01` `=` `0` Move the constant to the right`(u+0.9)^23.01` `=` `0` `(u+0.9)^23.01` `+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)` `=` `0.9sqrt(3.01)` `0.9+sqrt(3.01)`
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