Word problem

Question

Find two positive real numbers that differ by 2 and have a product of 10

Answer 1

x and x+2

x(x+2)=10

x^2+2x=10

x^2+2x-10=0

using the quadratic equation...

(-2+√4+40)/2

(-2+√44)/2

(-2+2√11)/2

-1+√11

use -1+√11 for x then x+2 will be 1+√11

the answers are √11+1 and √11-1

(√11+1)-(√11-1)=√11+1-√11+1=2

(√11+1)(√11-1)=11-1=10

 

Answer 2

Assign variables for the quantities we want to find.

 

Let x = one number

Let y = the second number

 

 

Set equations using the variables and info given.

 

x - y = 2             eq1

 

xy = 10              eq2

 

 

Substitute eq2 into eq1, so that eq1 in terms of x.

 

x - (10 / x) = 2

 

 

The LCD of the equation is x. 

 

(x² - 10) / x = 2x / x

 

 

Since the denominators on both sides of equation are equal, we only need to equate the numerators.

 

x² - 10 = 2x

 

 

Subtract 2x on both sides of equation.

 

x² - 2x - 10 = 0

 

 

Use the quadratic formula to solve for x.

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 1

b = -2

c = -10

 

x = (2 ± √(4 - 4(-10))) / 2

x = (2 ± √(4 + 40)) / 2

x = (2 ± √(44)) / 2

x = (2 ± 6.63) / 2

 

x =  (2 + 6.63) / 2                and              x = (2 - 6.63) / 2

x = 4.315                            and               x = -2.315

 

 

Accept the positive value of x:  x = 4.315

 

Substitute this value of x into eq2.

 

 

y = 10 / x

y = 10 / 4.315

y = 2.317

 

 

The two numbers are 4.315 and 2.317

Answer 3