Quadratics Word Problems 1

Question 1 of 5

The throw of a javelin thrower is modeled as

y = - 0.0189 x^{2} + x + 6.3

$y=-0.0189x^2+x+6.3$ , where

x

$x$ is the distance of the throw and

y

$y$ 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{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

To find the distance of the throw, equate the function to

0

$0$ and solve for

x

$x$ using the Quadratic Formula

First, list the coefficients of the quadratic equation individually

- 0.0189 x^{2} + x + 6.3 = 0

$-0.0189x^2+x+6.3=0$

$a = - 0.0189$ $a=-0.0189$ $b = 1$ $b=1$ $c = 6.3$ $c=6.3$

Substitute the values into the Quadratic Formula

$x$ $x$	$=$ $=$	$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	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 $a, b$ $a, b$ and $c$ $c$

	$=$ $=$	$\frac {-1 \pm \sqrt {1 +0.47628} }{-0.0378}$

	$=$ $=$	$\frac {-1 \pm \sqrt {1.47628} }{-0.0378}$

	$=$ $=$	$\frac {-1 \pm 1.215 }{-0.0378}$

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

$58.598$ m

58.598

$58.598$ m

Question 2 of 5

Find

x

$x$ given that the area of the rectangle below is

144

$144$

Round your answer to three decimal places

$x =$ $x=$ (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 =

$A=$ breadth

\times

$xx$ length

Substitute the given values into the Area Formula, then solve for

x

$x$ by Completing the Square

First, slot the given values into the Area Formula to form a quadratic equation

A = 144

$A=144$

breadth

= x

$=x$

length

= x + 12

$=x+12$

$A$ $A$	$=$ $=$	breadth $\times$ $xx$ length	Area of a Rectangle
$144$ $144$	$=$ $=$	$x (x + 12)$ $x(x+12)$	Substitute values
$x (x + 12)$ $x(x+12)$	$=$ $=$	$144$ $144$
$x^{2} + 12 x$ $x^2+12x$	$=$ $=$	$144$ $144$

Proceed with taking the coefficient of the

x

$x$ term, dividing it by two and then squaring it.

$x^{2} +$ $x^2+$ $12$ $12$ $x$ $x$	$=$ $=$	$144$ $144$			Coefficient of the $x$ $x$ term

$12$ $12$ $\div 2$ $-:2$	$=$ $=$	$6$ $6$			Divide it by $2$ $2$

		${(6)}^{2}$ $(6)^2$	$=$ $=$	$36$ $36$	Square

This number will make the left side a perfect square.

Add $36$ $36$ to both sides of the equation to keep the balance.

$x^{2} + 12 x$ $x^2+12x$	$=$ $=$	$144$ $144$
$x^{2} + 12 x$ $x^2+12x$ $+ 36$ $+36$	$=$ $=$	$144$ $144$ $+ 36$ $+36$	Add $36$ $36$ to both sides
$x^{2} + 12 x + 36$ $x^2+12x+36$	$=$ $=$	$180$ $180$

Now, transform the left side into a square of a binomial by factoring or using cross method.

$(x + 6) (x + 6)$ $(x+6)(x+6)$	$=$ $=$	$180$ $180$
$(x+6)^2$	$=$	$180$

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,

$x=7.416$

Question 3 of 5

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

Question 4 of 5

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

$36\text(cm)^2$

1. Not possible
2. Possible

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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 solutions

This means it is not possible for the triangle to have an area of

$36\text(cm)^2$

Not possible

Question 5 of 5

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 day

Substitute

$x=30$ 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$

$V$	$=$	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 $4$
$(x-8)^2$	$=$	$72$

Quadratics Word Problems 1

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Quiz summary

Information

1. Question

Hint

Well Done!

The Quadratic Formula

2. Question

Hint

Excellent!

Area of a Rectangle

3. Question

Hint

Keep Going!

Volume of a Rectangular Prism

4. Question

Hint

Great Work!

Discriminant Formula

Area of a Triangle

5. Question

Hint

Fantastic!

Axis of Symmetry

Quizzes

$(x-8)^2$	$=$	$72$
$sqrt((x-8)^2)$	$=$	$sqrt72$	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

	$=$	$8+6sqrt2$
	$=$	$16.485$

	$=$	$8-6sqrt2$
	$=$	$-0.485$