Last modified on March 28th, 2023

Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:

- By Completing the Square
- By Factoring
- By Quadratic Formula
- By graphing

For each process, follow the following typical steps:

- Make the equation
- Solve for the unknown variable using the appropriate method
- Interpret the result

**The product of two consecutive integers is 462. Find the numbers?**

Solution:

Let the numbers be x and x + 1

According to the problem,

x(x + 1) = 483

=> x^{2} + x – 483 = 0

=> x^{2} + 22x – 21x – 483 = 0

=> x(x + 22) – 21(x + 22) = 0

=> (x + 22)(x – 21) = 0

=> x + 22 = 0 or x – 21 = 0

=> x = {-22, 21}

Thus, the two consecutive numbers are 21 and 22.

**The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.**

Solution:

Let the numbers be n and n + 2

According to the problem,

=> n(n + 2) = 4[n + (n + 2)] – 1

=> n^{2} + 2n = 4[2n + 2] – 1

=> n^{2} + 2n = 8n + 7

=> n^{2} – 6n – 7 = 0

=> n^{2} -7n + n – 7 = 0

=> n(n – 7) + 1(n – 7) = 0

=> (n – 7) (n – 1) = 0

=> n – 7 = 0 or n – 1 = 0

=> n = {7, 1}

If n = 7, then n + 2 = 9

If n = 1, then n + 1 = 2

Since 1 and 2 are not possible. The two numbers are 7 and 9

**A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t ^{2} + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.**

Solution:

Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola.

Given the function is in the standard form h(t) = a^{2}x + bx + c, the formula to calculate the vertex is:

Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$

=> ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds

Thus, the time the projectile takes to reach the highest point is 2 seconds

${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2)^{2} – 64(2) + 80 = 144 feet

Thus, the maximum height the projectile reaches is 144 feet

**The difference between the squares of two consecutive even integers is 68. Find the numbers.**

Solution:

Let the numbers be x and x + 2

According to the problem,

(x + 2)^{2} – x^{2} = 68

=> x^{2} + 4x + 4 – x^{2} = 68

=> 4x + 4 = 68

=> 4x = 68 – 4

=> 4x = 64

=> x = 16

Thus the two numbers are 16 and 18

**The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.**

Solution:

Let the number be x

Thus,

Length = 2x + 5

Breadth = x – 4

According to the problem,

(2x + 5)(x – 4) = 15

=> 2x^{2} – 8x + 5x – 20 – 15 = 0

=> 2x^{2} – 3x – 35 = 0

=> 2x^{2} – 10x + 7x – 35 = 0

=> 2x(x – 5) + 7(x – 5) = 0

=> (x – 5)(2x + 7) = 0

=> x – 5 = 0 or 2x + 7 = 0

=> x = {5, -7/2}

Since we cannot have a negative measurement in mensuration, the number is 5 inches.

Now,

Length = 2x + 5 = 2(5) + 5 = 15 inches

Breadth = x – 4 = 15 – 4 = 11 inches

**A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².**

Solution:

Given,

Length of the garden = 50 cm

Width of the garden = 34 cm

Let the uniform width of the boundary be = x cm

According to the problem,

(50 + 2x)(34 + 2x) – 50 × 34 = 540

=> 4x^{2} + 168x – 540 = 0

=> x^{2} + 42x – 135 = 0

Since, this quadratic equation is in the standard form ax^{2} + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135

x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

=> ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$

=> ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$

=> ${\dfrac{-42\pm \sqrt{2304}}{2}}$

=> ${\dfrac{-42\pm 48}{2}}$

=> ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$

=> x = {-45, 3}

Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm

**The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.**

Solution:

Let the length of the other two sides be x and x + 4

According to the problem,

(x + 4)^{2 }+ x^{2 }= 20^{2}

=> x^{2} + 8x + 16 + x^{2} = 400

=> 2x^{2} + 8x + 16 = 400

=> 2x^{2} + 8x – 384 = 0

=> x^{2} + 4x – 192 = 0

=> x^{2} + 16x – 12x – 192 = 0

=> x(x + 16) – 12(x + 16) = 0

=> (x + 16)(x – 12) = 0

=> x + 16 = 0 and x – 12 = 0

=> x = {-16, 12}

Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16

**Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t ^{2} +16t + 480, where t is the time in seconds and h is the height in feet.**

Solution:

Comparing the given function with the given function f(x) = ax^{2} + bx + c, here a = -16, b = 16, c = 480

a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height

x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds

Thus Jennifer took 0.5 seconds to reach her maximum height

b) Putting the value of the vertex by substitution in the function, we get

${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$

=> ${-16\left( \dfrac{1}{4}\right) +8+480}$

=> 484 feet

Thus the highest point that Jennifer reached was 484 feet

c) When Jennifer hit the water, her height was 0

Thus, by substituting the value of the height in the function, we get

-16t^{2} +16t + 480 = 0

=> -16(t^{2} + t – 30) = 0

=> t^{2} + t – 30 = 0

=> t^{2} + 6t – 5t – 30 = 0

=> t(t + 6) – 5(t + 6) = 0

=> (t + 6)(t – 5) = 0

=> t + 6 = 0 or t – 5 = 0

=> x = {-6, 5}

Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.

Last modified on March 28th, 2023