Lesson Find the number using quadratic equation

Find the number using quadratic equation

Problem 1

Two solutions are possible.


One solution is to guess the answer.

A person familiar with the multiplication table, can do it mentally (without using equations) in 5 seconds.



Alternative solution is algebraic.


    x + y = -3     (1)

    xy = -40.      (2)


From equation (1), express  y = -3 - x  and substitute to 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 factoring

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

OR using the quadratic formula

     =  =  = .


It gives you the answer that is placed at the very beginning of my post.

Problem 2

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


Then you write the condition for the product


    x*(x+6) = 616,

or

    x^2 + 6x - 616 = 0.


Use the quadratic formula to solve this quadratic equation.


     =  =  = 


Only the positive root is the solution to the problem  x =  =  = 22.


ANSWER.  The numbers are 22 and 28.

Problem 3

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

 =  = ,

 =  = 17,   = 11.

 =  = -11,   = -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

After reading the condition, turn on your mind.


      = 3  = 3 + .


So, the number is  3  and its reciprocal is  .    


It is one answer.  Another answer is THIS:


    The number is    and its reciprocal is 3.

Let x be the number; then reciprocal is  .

The equation is THIS:

    x +  = .


Multiply both sides by 3x


    3x^2 + 3 = 10x

    3x^2 - 10x + 3 = 0

     =  =  = .


One root is   =  =  = 3.


The other root is   =  =  = .


Thus we solved the problem formally and obtained  BOTH  solutions (!)

Problem 5

Let x be the greater number, y be the lesser number.
Then from the condition you have this system of two equations

x - y = 10,      (1)

 +  = .     (2)


From equation (1), express x = 10 + y  and substitute it into equation (2), replacing x.

You will get a single equation for y:

 +  = .


To solve it, multiply both sides by 4y*(10+y).  You will get

12y + 4*(10+y) = y*(10+y).


Simplify and solve for y:


12y + 40 + 4y = y^2 + 10y,

y^2 - 6y - 40 = 0.


Factor left side

(y-10)*(y+4) = 0.


The last equation has two solutions y= 10  and  y= -4,  but only one of them, y= 10, is positive.

The other solution does not work.


So, the problem has only one solution:  y= 10,  x = y+10 = 10+10 = 20.


Answer.  The solution is  x= 20, y= 10.

Check.   x - y = 10;

         +  =  +  =  =  =    ! Correct !

Problem 6

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                    = +/-  = +/- 18.


So, the positive odd integers are 17 and 19;

    the negative odd integers are -19 and -17.