Quadratics Word Problems 2
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Question 1 of 5
1. Question
The measurements of a parabolic tunnel and a freight train is shown below. Would the train fit in the tunnel?
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The Quadratic Formula
$$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$$Form the quadratic equation of the parabola from the tunnel and graph it together with the measurements of the train, to see if the train can be drawn inside the parabola.First, turn the tunnel into a graph, using `x` axis as the ground and `y` axis as the axis of symmetry
Proceed with forming the equation of the parabolaStart by identifying the `x` and `y` intercepts from the graph`p` `=` `-4` `x`-intercept `q` `=` `4` `x`-intercept `y` `=` `10` `y`-intercept `x` `=` `0` Value of `x` at the `y`-intercept Now, slot these values into the Intercept Form and solve for `a``y` `=` `a(x-p)(x-q)` Intercept Form `10` `=` `a(0+4)(0-4)` Substitute values `10` `=` `a(4)(-4)` `10` `=` `-16a` `10``-:(-16)` `=` `-16a``-:(-16)` `-5/8` `=` `a` `a` `=` `-5/8` Substitute `a` and the `x` intercepts into the Intercept Form`y` `=` `a(x-p)(x-q)` Intercept Form `y` `=` `-5/8(x-4)(x-4)` Substitute values `y` `=` `-5/8(x^2-16)` This is the equation for the parabolaThis time, turn the train into a graphSimilar to the tunnel, use `x` axis as the ground and `y` axis as the axis of symmetry
Substitute `x=2.6` to the equation of the parabola and solve for `y`If `y` is greater that `5.9` (height of the train), the train will fit in the tunnel`y` `=` `-5/8(x^2-16)` Equation of the Parabola `y` `=` `-5/8(2.6^2-16)` Substitute `x=2.6` `y` `=` `-5/8(6.76-16)` `y` `=` `-5/8(-9.24)` `y` `=` `5.775` `5.775` is the height of the tunnel at `x=2.6`Since `5.775``<``5.9`, the train will not fit through the tunnelNo -
Question 2 of 5
2. Question
Find `x` given that the area of the rectangle below is `36`
Round your answer to three decimal places- `x=` (4.325)
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Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.Area of a Rectangle
`A=`breadth`xx`lengthSubstitute the given values into the Area Formula, then solve for `x` by Completing the SquareFirst, slot the given values into the Area Formula to form a quadratic equation`A=36`breadth`=x`length`=x+4``A` `=` breadth`xx`length Area of a Rectangle `36` `=` `x(x+4)` Substitute values `x(x+4)` `=` `36` `x^2+4x` `=` `36` Proceed with taking the coefficient of the `x` term, dividing it by two and then squaring it.`x^2+``4``x` `=` `36` Coefficient of the `x` term `4``-:2` `=` `2` Divide it by `2` `(2)^2` `=` `4` Square This number will make the left side a perfect square.Add `4` to both sides of the equation to keep the balance.`x^2+4x` `=` `36` `x^2+4x` `+4` `=` `36` `+4` Add `4` to both sides `x^2+4x+4` `=` `40` Now, transform the left side into a square of a binomial by factoring or using cross method.
`(x+2)(x+2)` `=` `40` `(x+2)^2` `=` `40` Finally, take the square root of both sides and continue solving for `x`.`(x+2)^2` `=` `40` `sqrt((x+2)^2)` `=` `sqrt40` Take the square root `x+2` `=` `+-2sqrt10` Square rooting a number gives a plus and minus solution `x+2` `-2` `=` `+-2sqrt10` `-2` Subtract `2` from both sides `x` `=` `-2+-2sqrt10` Simplify The roots can also be written individually`=` `-2+2sqrt10` `=` `4.325` `=` `-2-2sqrt10` `=` `-8.325` Length cannot be a negative value. Therefore, `x=4.325``x=4.325` -
Question 3 of 5
3. Question
Find `x` given the measurements below
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The cross method is a factorisation method used for quadratics.Equate the given equation to the given area`A=6x^2+16x``A=32``6x^2+16x` `=` `32` `6x^2+16x` `-32` `=` `32` `-32` Subtract `32` from both sides `6x^2+16x-32` `=` `0` `(6x^2+16x-32)``-:2` `=` `0``-:2` Divide both sides by `2` `3x^2+8x-16` `=` `0` Factorise the equation using the cross method
`(3x-4)(x+4)` `=` `0` Compute for `x` separately`3x-4` `=` `0` `3x-4` `+4` `=` `0` `+4` `3x` `=` `4` `3x``-:3` `=` `4``-:3` `x` `=` `4/3` `x+4` `=` `0` `x+4` `-4` `=` `0` `-4` `x` `=` `-4` Length cannot be a negative value. Therefore, `x=4/3``x=4/3` -
Question 4 of 5
4. Question
Find `x` given that the area of the triangle below is `45`
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Area of a Triangle
`A=1/2bh`Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.Substitute the given values to the Area Formula to form a quadratic equation, then solve for `x` by completing the squareFirst, slot the given values into the Area Formula to form a quadratic equation`A=45`base`=x+6`height`=x``A` `=` `1/2bh` Area of a Triangle `45` `=` `1/2(x+6)(x)` Substitute values `45``xx2` `=` `1/2(x+6)(x)``xx2` Multiply both sides to `2` `90` `=` `(x+6)x` `90` `=` `x^2+6x` `x^2+6x` `=` `90` Proceed with taking the coefficient of the `x` term, dividing it by two and then squaring it.`x^2+``6``x` `=` `90` Coefficient of the `x` term `6``-:2` `=` `3` Divide it by `2` `(3)^2` `=` `9` Square This number will make the left side a perfect square.Add `9` to both sides of the equation to keep the balance.`x^2+6x` `=` `90` `x^2+6x` `+9` `=` `90` `+9` Add `9` to both sides `x^2+6x+9` `=` `99` `(x+3)^2` `=` `99` Finally, take the square root of both sides and continue solving for `x`.`(x+3)^2` `=` `99` `sqrt((x+3)^2)` `=` `sqrt99` Take the square root `x+3` `=` `+-3sqrt11` Square rooting a number gives a plus and minus solution `x+3` `-3` `=` `+-3sqrt11` `-3` Subtract `3` from both sides `x` `=` `-3+-3sqrt11` Simplify The roots can also be written individually`=` `-3+3sqrt11` `=` `6.9499` `=` `-3-3sqrt11` `=` `-12.9499` Length cannot be a negative value. Therefore, `x=6.9499``x=6.9499` -
Question 5 of 5
5. Question
Using only `60`m of fencing, find the maximum possible area for a paddock. This paddock is against a barn, which means only `3` sides need to be fenced.
- (450)`\text(m)^2`
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Axis of Symmetry
$$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$Area of a Rectangle
`A=`breadth`xx`lengthAssign a variable for one side of the paddock and create a quadratic equation using the Area formula. Then, find the maximum area with the help of the axis of symmetry.First, assign a variable to each side of the rectangle that needs fencing
The two equal sides can be labelled as `x`. Since the fence will be `60`m long in total, the last side would be `60-2x`Form a quadratic equation by substituting values to the Area Formulabreath`=x`length`=60-2x``A` `=` breadth`xx`length Area Formula `A` `=` `x(60-2x)` Substitute values `A` `=` `60x-2x^2` `A` `=` `-2x^2+60x` Next, solve for the axis of symmetry of the equation`A=-2x^2+60x``a=-2` `b=60` `c=0``x` `=` $$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$ Axis of Symmetry `x` `=` $$\frac{-\color{#9a00c7}{60}}{2\color{#00880A}{(-2)}}$$ Substitute values `x` `=` `(-60)/(-4)` `x` `=` `15` This means that the side `x` should be `15`m long for the paddock to have the maximum areaFinally, substitute `x=15` to the given equation to get the maximum area`A` `=` `-2x^2+60x` `=` `-2(15)^2+60(15)` Substitute `x=15` `=` `-2(225)+900` `=` `-450+900` `=` `450` Therefore, the maximum area is `450\text(m)^2``450\text(m)^2`
Quizzes
- Sum & Product of Roots 1
- 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
- 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
- Solving Reducible Equations