WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS ON NUMBER

Example 1 :

The product of the number and the number increased by 4 is 96. Find the two possible answers for the number.

Solution :

Let the smallest number  =  x.

Largest number  =  x + 4

(x) (x + 4)  =  96

x² + 4x – 96  =  0

By factorization, we get

(x + 12) (x – 8)  =  0

x + 12  =  0 or x – 8  =  0

x  =  - 12 or x  =  8

So, the possible numbers are 8 or - 12

Example 2 :

When 24 is subtracted from the square of a number the result is five times the original number. Find the number ?

Solution :

Let the number  =  x.

Square of number  =  x²

x² – 24  =  5x

x² – 5x – 24  =  0

By factorization, we get

(x – 8) (x + 3)  =  0

x – 8  =  0 or x + 3  =  0

x  =  8 or x  =  - 3

So, the numbers are 8 or – 3.

Example 3 :

The sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are 

Solution :

Let x and y be two numbers.

Sum of two numbers  =  8

x+y  =  8  ------(1)

Sum of their squares  =  34

x²+y²  =  34 ------(2)

From (1), y = 8-x

x²+(8-x)²  =  34

x² + 64 + x² - 16x  =  34

2x²-16x+64-34  =  0

2x²-16x+30  =  0

Dividing it by 2, we get

x²-8x+15  =  0

(x-3)(x-5)  =  0

x  =  3 and x  =  5

So, the numbers are 3 and 5.

Example 4 :

The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are,

Solution :

Let x be the smaller number and x+3 be two integers.

Sum of their squares  =  89

x²+(x+3)²  =  89

x²+x²+6x+9-89  =  0

2x²+6x-80  =  0

x²+3x-40  =  0

(x+8)(x-5)  =  0

Solving it, we get

x  =  -8 and x  =  5

x+3  ==>  5+3  ==> 8

So, the integers are 5 and 8.

Example 5 :

There are two consecutive numbers such that the difference of their reciprocals is 1/240. The numbers are 

Solution :

Let two consecutive numbers are x and x+1.

Difference of their reciprocals  =  1/240

(1/(x+1)) - (1/x)  =  1/240

(x-(x+1))/x(x+1)  =  1/240

x-x-1/(x²+x)  =  1/240

-1/(x²+x)  =  1/240

-240  =  x²+x

x²+x-240  =  0

(x+16)(x-15)  =  0

Equating each factor to zero, we get

x  =  -16 and x  =  15.

So, the consecutive numbers are 15 and 16.

Example 6 :

The sum of two irrational numbers multiplied by the larger one is 70 and their difference is multiplied by the smaller one is 12, the two numbers are

Solution :

Let two irrational numbers are √a and √b.

Sum of two irrational numbers multiplied by larger  =  70

√a(√a + √b)  =  70

a+√ab  =  70  -----(1)

From (1)

√ab  =  70-a

√b(√a - √b)  =  12

√ab-b  =  12  -----(2)

By applying the value of √ab  =  70-a in (2), we get

(70-a)-b  =  12

70-a-b  =  12

a+b  =  58

b  =  58-a

√a(58-a)-(58-a)  =  12 

√a(58-a)  =  12+58-a

√58a-a²  =  70-a

Taking squares on both sides, we get

58a-a²  =  (70-a)²

58a-a²  =  4900+a² - 140a

2a² -140a-58a+4900  =  0

2a² -198a+4900  =  0

a² -99a+2450  =  0

(a-50)(a-49)  =  0

a  =  49 and x  =  50

So, the numbers are 50 and 49.

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