Quadratics Word Problems 2

Question 1 of 5

The measurements of a parabolic tunnel and a freight train is shown below. Would the train fit in the tunnel?

1. Yes
2. No

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The Quadratic Formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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

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

$x$ axis as the ground and

y

$y$ axis as the axis of symmetry

Proceed with forming the equation of the parabola

Start by identifying the

x

$x$ and

y

$y$ intercepts from the graph

$p$ $p$	$=$ $=$	$- 4$ $-4$	$x$ $x$ -intercept
$q$ $q$	$=$ $=$	$4$ $4$	$x$ $x$ -intercept
$y$ $y$	$=$ $=$	$10$ $10$	$y$ $y$ -intercept
$x$ $x$	$=$ $=$	$0$ $0$	Value of $x$ $x$ at the $y$ $y$ -intercept

Now, slot these values into the Intercept Form and solve for

a

$a$

$y$ $y$	$=$ $=$	$a (x - p) (x - q)$ $a(x-p)(x-q)$	Intercept Form
$10$ $10$	$=$ $=$	$a (0 + 4) (0 - 4)$ $a(0+4)(0-4)$	Substitute values
$10$ $10$	$=$ $=$	$a (4) (- 4)$ $a(4)(-4)$
$10$ $10$	$=$ $=$	$- 16 a$ $-16a$
$10$ $10$ $\div (- 16)$ $-:(-16)$	$=$ $=$	$- 16 a$ $-16a$ $\div (- 16)$ $-:(-16)$

$- \frac{5}{8}$ $-5/8$	$=$ $=$	$a$ $a$

$a$ $a$	$=$ $=$	$- \frac{5}{8}$ $-5/8$

Substitute

a

$a$ and the

x

$x$ intercepts into the Intercept Form

$y$ $y$	$=$ $=$	$a (x - p) (x - q)$ $a(x-p)(x-q)$	Intercept Form

$y$ $y$	$=$ $=$	$- \frac{5}{8} (x - 4) (x - 4)$ $-5/8(x-4)(x-4)$	Substitute values

$y$ $y$	$=$ $=$	$- \frac{5}{8} (x^{2} - 16)$ $-5/8(x^2-16)$

This is the equation for the parabola

This time, turn the train into a graph

Similar to the tunnel, use

x

$x$ axis as the ground and

y

$y$ axis as the axis of symmetry

Substitute

x = 2.6

$x=2.6$ to the equation of the parabola and solve for

y

$y$

If

y

$y$ is greater that

5.9

$5.9$ (height of the train), the train will fit in the tunnel

$y$ $y$	$=$ $=$	$- \frac{5}{8} (x^{2} - 16)$ $-5/8(x^2-16)$	Equation of the Parabola

$y$ $y$	$=$ $=$	$- \frac{5}{8} ({2.6}^{2} - 16)$ $-5/8(2.6^2-16)$	Substitute $x = 2.6$ $x=2.6$

$y$ $y$	$=$ $=$	$-5/8(6.76-16)$

$y$	$=$	$-5/8(-9.24)$

$y$	$=$	$5.775$

$5.775$ is the height of the tunnel at

$x=2.6$

Since

$5.775$

$<$

$5.9$ , the train will not fit through the tunnel

No

Question 2 of 5

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

$xx$ length

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

$x$ by Completing the Square

First, 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,

$x=4.325$

Question 3 of 5

Find

$x$ given the measurements below

1. $4/3$
2. $6$
3. $4$
4. $3/4$

Question 4 of 5

Find

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

$45$

1. $6.9499$
2. $6.9499$ and $-12.9499$
3. $5.3499$
4. $5.3499$ and $-8.3499$

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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 square

First, 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,

$x=6.9499$

Question 5 of 5

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

$xx$ length

Assign 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 Formula

breath

$=x$

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 area

Finally, 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

$450\text(m)^2$

$6x^2+16x$	$=$	$32$
$6x^2+16x$ $-32$	$=$	$32$ $-32$	Subtract $32$ from both sides
$6x^2+16x-32$	$=$	$0$
$(6x^2+16x-32)$ $-:2$	$=$	$0$ $-:2$	Divide both sides by $2$
$3x^2+8x-16$	$=$	$0$

Quadratics Word Problems 2

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Quiz summary

Information

1. Question

Hint

Nice Job!

The Quadratic Formula

2. Question

Hint

Fantastic!

Area of a Rectangle

3. Question

Hint

Correct!

4. Question

Hint

Nice Job!

Area of a Triangle

5. Question

Hint

Correct!

Axis of Symmetry

Area of a Rectangle

Quizzes

$3x-4$	$=$	$0$
$3x-4$ $+4$	$=$	$0$ $+4$
$3x$	$=$	$4$
$3x$ $-:3$	$=$	$4$ $-:3$

$x$	$=$	$4/3$