Lesson Find a number using a quadratic equation

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> Lesson Find a number using a quadratic equation      Log On


   

This Lesson (Find a number using a quadratic equation) was created by by ikleyn(52748) About Me : View Source, Show
About ikleyn:

Find a number using a quadratic equation


Problem 1

Find two integers with a product of  -40  and a sum of  -3.

Solution

ANSWER.   The numbers are  -8  and  5. 

Two solutions are possible.


One solution is to guess the answer mentally.

A person familiar with the multiplication table, can do it mentally (without using equations) in 10 seconds.



Alternative solution is algebraic. Write equations as you read the problem

    x + y = -3     (1)

    xy = -40.      (2)


From equation (1), express  y = -3 - x  and substitute into equation (2)

    x*(-3-x) = -40.


Simplify it and reduce to the standard form of the quadratic equation

    -x^2 - 3x + 40 = 0,   or, equivalently,

     x^2 + 3x - 40 = 0.


You may solve the last equation either by factoring

    (x+8)*(x-5) = 0

or using the quadratic formula

    x%5B1%2C2%5D = %28-3+%2B-+sqrt%283%5E2+-+4%2A%28-40%29%29%29%2F2 = %28-3+%2B-+sqrt%28169%29%29%2F2 = %28-3+%2B-+13%29%2F2.


It gives you the same answer that is placed at the very beginning.

Problem 2

Two positive integers differ by  6.  Their product is  616.  Find the integers.

Solution 

Let x be the smaller integer; then the greater integer is (x+6).


Then you write this equation for the product


    x*(x+6) = 616,

or

    x^2 + 6x - 616 = 0.


Use the quadratic formula to solve this quadratic equation.


    x%5B1%2C2%5D = %28-6+%2B-+sqrt%28%28-6%29%5E2+%2B+4%2A616%29%29%2F2 = %28-6+%2B-+sqrt%282500%29%29%2F2 = %28-6+%2B-+50%29%2F2


Only positive root is the solution to the problem  x = %28-6+%2B+50%29%2F2 = 44%2F2 = 22.


ANSWER.  The numbers are 22 and 28.

Problem 3

The difference of two positive numbers is six.  Their product is  223  less than the sum of their squares.
What are the two numbers?

Solution 

Let x be the larger number, y be the lesser number.


Then

x - y = 6,                (1)
xy = x^2 + y^2 - 223.     (2)


Square equation (1) (both sides).  Keep equation (2) as is:

x^2 - 2xy + y^2 =  36,    (3)
x^2 - xy  + y^2 = 223     (4)    (<<<---=== it is transformed eq(2) )

----------------------------------Subtract eq(3) from eq(4). You will get

      xy        = 187.


Now you have system of two equations

x - y =   6,
xy    = 187.


It is reduced to the quadratic equation

x*(x-6) = 187

x^2 -6x - 187 = 0

x%5B1%2C2%5D = %286+%2B-+sqrt%2836+%2B+4%2A187%29%29%2F2 = %286+%2B-+28%29%2F2,

x%5B1%5D = %286%2B28%29%2F2 = 17,  y%5B1%5D = 11.

x%5B2%5D = %286-28%29%2F2 = -11,  y%5B2%5D = -17.


Answer.  There are TWO solutions:   a) (x,y) = (17,11);  b) (x,y) = (-11,-17).      

         Since the problem asks about positive numbers, only first pair satisfies this requirement.

Problem 4

The sum of a number and its reciprocal is  10/3.  What is the number?

Solution

There are two ways to solve this problem. One way is easy and gives the solution in 3-4-5 seconds.

The other way is formal, through solving a quadratic equation.

I will show you both ways.


                Easy way 


After reading the condition, turn on your mind.


     10%2F3 = 3 1%2F3 = 3 + 1%2F3.


So, the number is  3  and its reciprocal is  1%2F3.    


It is one answer.  Another answer is THIS:


    The number is  1%2F3  and its reciprocal is 3.

Thus easy way is complete.


                The formal way 


Let x be the number; then reciprocal is  1%2Fx.

The equation is THIS:

    x + 1%2Fx = 10%2F3.


Multiply both sides by 3x


    3x^2 + 3 = 10x

    3x^2 - 10x + 3 = 0

    x%5B1%2C2%5D = %2810+%2B-+sqrt%2810%5E2+-+4%2A3%2A3%29%29%2F%282%2A3%29 = %2810+%2B-+sqrt%2864%29%29%2F6 = %2810+%2B-+8%29%2F6.


One root is  x%5B1%5D = %2810%2B8%29%2F6 = 18%2F6 = 3.


The other root is  x%5B2%5D = %2810-8%29%2F6 = 2%2F6 = 1%2F3.


Thus we solved the problem formally and mentally (!)


For advanced young students,  who are able manipulate fractions freely,
it is an  entertainment problem,  which  SHOULD  be solved mentally,  and it is fan for them to do it.


For other students,  the formal way is often very boring and torturing way to solve the problem formally.


Good student should know both ways.

Therefore,  I placed both ways solutions here with the easy  joking  mode first.


Problem 5

The product of two consecutive odd integers is  323.  Find the integers.

Solution 

Let x be the EVEN integer exactly midway between the two odd consecutive integers.


Then the odd integers are (x-1) and (x+1), and


    (x+1)*(x-1) = 323.   or


     x^2 - 1    = 323

     x^2        = 323 + 1 = 324

     x                    = +/- sqrt%28324%29 = +/- 18.


So, the positive odd integers are 17 and 19;

    the negative odd integers are -19 and -17.

Problem 6

A positive real number is  2  less than another.
If the sum of the squares of the two numbers is  6,  find the numbers.

Solution 

Let x be the smaller positive real number.


Then the larger number is (x+2), and the problem leads to equation

    x^2 + (x+2)^2 = 6    (the sum of squares is 6).


Simplify and find x

    x^2 + (x^2 + 4x + 4) = 6

    2x^2 + 4x -2 = 0

     x^2 + 2x - 1 = 0


Use the quadratic formula

     x%5B1%2C2%5D = %28%28-2%29+%2B-+sqrt%282%5E2+-+4%2A1%2A%28-1%29%29%29%2F2 = %28-2+%2B-+sqrt%288%29%29%2F2 = -1+%2B-+sqrt%282%29.


ANSWER.  The numbers are  x = -1%2Bsqrt%282%29  (the positive root) and -1-sqrt%282%29.


My other lessons on solving quadratic equations in this site are
    - HOW TO solve quadratic equation by completing the square - Learning by examples
    - Solving quadratic equations without quadratic formula
    - Who is who in quadratic equations
    - Using Vieta's theorem to solve quadratic equations and related problems

    - Using quadratic equations to solve word problems
    - Challenging word problems solved using quadratic equations
    - Business-related problems to solve them using quadratic equations
    - Rare beauty investment problem to solve using quadratic equation
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations
    - Entertainment problems on quadratic equations
    - OVERVIEW of lessons on solving quadratic equations

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

This lesson has been accessed 1016 times.