Completing the Square 2
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Question 1 of 5
1. Question
Solve by completing the square`2x^2-12x=-8`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.
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
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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.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
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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.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
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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.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
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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)`
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