Quadratics Word Problems 1
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Question 1 of 5
1. Question
The throw of a javelin thrower is modeled as `y=-0.0189x^2+x+6.3`, where `x` is the distance of the throw and `y` is the height of the throw. Find the value of the distance of the throw in metres.Round your answer to three decimal places- (58.598)m
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The Quadratic Formula
$$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$$To find the distance of the throw, equate the function to `0` and solve for `x` using the Quadratic FormulaFirst, list the coefficients of the quadratic equation individually`-0.0189x^2+x+6.3=0``a=-0.0189` `b=1` `c=6.3`Substitute the values into the Quadratic Formula`x` `=` $$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$$ Quadratic Formula `=` $$\frac {- \color{#9a00c7}{1} \pm \sqrt {\color{#9a00c7}{1}^2-4\color{#00880A}{(-0.0189)}\color{#007DDC}{(6.3)}} }{2 \color{#00880A}{(-0.0189)}}$$ Plug in the values of `a, b` and `c` `=` $$\frac {-1 \pm \sqrt {1 +0.47628} }{-0.0378}$$ `=` $$\frac {-1 \pm \sqrt {1.47628} }{-0.0378}$$ `=` $$\frac {-1 \pm 1.215 }{-0.0378}$$ Write each root individually$$x_1$$ `=` $$\frac {-1+1.215 }{-0.0378}$$ `=` $$\frac {0.215}{-0.0378}$$ `=` $$-5.688$$ $$x_2$$ `=` $$\frac {-1-1.215 }{-0.0378}$$ `=` $$\frac {-2.215}{-0.0378}$$ `=` $$58.598$$ Distance cannot be a negative value. Therefore, the distance of the throw is `58.598`m`58.598`m -
Question 2 of 5
2. Question
Find `x` given that the area of the rectangle below is `144`
Round your answer to three decimal places- `x=` (7.416)
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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=144`breadth`=x`length`=x+12``A` `=` breadth`xx`length Area of a Rectangle `144` `=` `x(x+12)` Substitute values `x(x+12)` `=` `144` `x^2+12x` `=` `144` Proceed with taking the coefficient of the `x` term, dividing it by two and then squaring it.`x^2+``12``x` `=` `144` Coefficient of the `x` 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` `=` `144` `x^2+12x` `+36` `=` `144` `+36` Add `36` to both sides `x^2+12x+36` `=` `180` Now, transform the left side into a square of a binomial by factoring or using cross method.
`(x+6)(x+6)` `=` `180` `(x+6)^2` `=` `180` Finally, take the square root of both sides and continue solving for `x`.`(x+6)^2` `=` `180` `sqrt((x+6)^2)` `=` `sqrt180` Take the square root `x+6` `=` `+-6sqrt5` Square rooting a number gives a plus and minus solution `x+6` `-6` `=` `+-6sqrt5` `-6` Subtract `6` from both sides `x` `=` `-6+-6sqrt5` Simplify The roots can also be written individually`=` `-6+6sqrt5` `=` `7.416` `=` `-6-6sqrt5` `=` `-19.416` Length cannot be a negative value. Therefore, `x=7.416``x=7.416` -
Question 3 of 5
3. Question
The corners of a square cardboard are cut so that the sides can be folded up to form an open box. Given the measurements below, find `x`.
Round your answer to three decimal places- `x=` (16.485)cm
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Volume of a Rectangular Prism
`V=`length`xx`breadth`xx`heightVisualize folding up the sides and note the values of the length, breadth, and height. Substitute these values to the Volume Formula and then solve for `x`First, list down of the measurements of the box when it is folded up
Note that `x-8` comes from `x-4-4`, or the side of the original square minus the two corners that were cutSubstitute the values to the Volume Formula, then solve for `x`Volume`=288`length`=x-8`breadth`=x-8`height`=4``V` `=` length`xx`breadth`xx`height Volume Formula `288` `=` `(x-8)(x-8)4` Substitute values `288` `=` `4(x-8)^2` `4(x-8)^2` `=` `288` `4(x-8)^2``-:4` `=` `288``-:4` Divide both sides by `4` `(x-8)^2` `=` `72` Take the square root of both sides and continue solving for `x`.`(x-8)^2` `=` `72` `sqrt((x-8)^2)` `=` `sqrt72` Take the square root `x-8` `=` `+-6sqrt2` Square rooting a number gives a plus and minus solution `x-8` `+8` `=` `+-6sqrt2` `+8` Add `8` to both sides `x` `=` `8+-6sqrt2` Simplify The roots can also be written individually`=` `8+6sqrt2` `=` `16.485` `=` `8-6sqrt2` `=` `-0.485` Length cannot be a negative value. Therefore, `x=16.485`cm`x=16.485`cm -
Question 4 of 5
4. Question
A wire is bent to form a right triangle as shown below. Find out if it is possible for the area of this triangle to be `36\text(cm)^2`
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Discriminant Formula
$$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$$Nature of the Roots Discriminant (`Delta`) Two real roots `Delta``>``0` One real root `Delta=0` No real roots `Delta``<``0` Area of a Triangle
`A=1/2bh`Substitute the given values to the Area Formula to form a quadratic equation. Then, solve for the discriminant of the equation to check if it would have solutions.First, slot the given values into the Area Formula to form a quadratic equation`A=36`base`=16-x`height`=x``A` `=` `1/2bh` Area of a Triangle `36` `=` `1/2(16-x)(x)` Substitute values `36``xx2` `=` `1/2(16-x)(x)``xx2` Multiply both sides to `2` `72` `=` `(16-x)x` `72` `=` `16x-x^2` `x^2-16x+72` `=` `0` Move all terms to the left Find the discriminant (`Delta`) of the quadratic equation to see if there would be solutions for `x``x^2-16x+72=0``a=1` `b=-16` `c=72``Delta` `=` $${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$$ Discriminant Formula `=` $${\color{#9a00c7}{-16}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(72)}$$ Substitute values `=` `256-288` `=` `-32` Since `Delta``<``0`, the equation has no real roots or solutionsThis means it is not possible for the triangle to have an area of `36\text(cm)^2`Not possible -
Question 5 of 5
5. Question
The profit in manufacturing `x` television panels per day is given by the profit relationship `P=-2x^2+120x-300`, where `p` is in dollars. Find the following:-
Number of panels for maximum profit: (30)Maximum profit per day: `$` (1500)
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Axis of Symmetry
$$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$Use the axis of symmetry to find the number of TV panels to be made for the maximum profit. Substitute this number back into the given equation to find the maximum profit.First, solve for the axis of symmetry`P=-2x^2+120x-300``a=-2` `b=120` `c=-300``x` `=` $$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$ Axis of Symmetry `x` `=` $$\frac{-\color{#9a00c7}{120}}{2\color{#00880A}{(-2)}}$$ Substitute values `x` `=` `(-120)/(-4)` `x` `=` `30` This means that the number of TV panels that will give the maximum profit is `30` TV panels per daySubstitute `x=30` to the given equation to get the maximum profit per day`P` `=` `-2x^2+120x-300` `=` `-2(30)^2+120(30)-300` Substitute `x=30` `=` `-2(900)+3600-300` `=` `-1800+3600-300` `=` `-1800+3600-300` `=` `1500` Therefore, the maximum profit is `$1500`Number of panels for maximum profit: `30`Maximum profit per day: `$1500` -
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