The sum of 2 positive integers is 151. The lesser number is 19 more than the square root of the greater number, What is the value of the greater number minus the lesser number?

### 6 Answers

x² + x + 19 = 151

x² + x – 132 = 0

x = 11 (valid) or -12 (invalid, as per the question)

The smaller number is 11+19 = 30 and the greater number is 121.

y = 151 – x

y = x^(1/2) + 19

151 – x = x^(1/2) + 19

x^(1/2) = 132 – x

x = x^2 – 264x + 132^2

x^2 – 265x + 17424 = 0

(x – 144)(x – 121) = 0

Solutions:

x = 144, y = 7

x = 121, y = 30

The numbers are 144 and 7 or 121 and 30

The value of the greater number minus

the lesser number is either 137 or 91.

A great way to start word problems is to write down everything that is in words in the problem in mathematical terms

– The sum of 2 positive integers is 151

(eq1)

a + b = 151

– The lesser number is 19 more than the square root of the greater number

(eq2)

a – 19 = squareroot(b)

now you can just plug in. Let’s set equation 2 to b= (I tried a= and it was pretty complicated)

b = (a-19)(a-19)

b = a^2 -38a + 361

and plug that into equation 1

a + (a^2 -38a + 361) = 151

set that equation = zero so that we can use the quadratic formula

a^2 – 37a + 210 = 0

I leave the quadratic formula to you. Remember that both a and b must be positive, so you should use the positive value, and not the negative value, that you get from the quadratic formula.

Just plug the value of a into equation 1 to get the value of b.

a + b^2 = 151

a = b + 19

b + 19 + b^2 = 151

b^2 + b – 132 = 0

b = (-1 +/- sqrt(1 + 528)) / 2

b = (-1 +/- sqrt(529)) / 2

b = (-1 +/- 23) / 2

b = -24/2 , 22/2

b = -12 , 11

a and b > 0

b = 11

b^2 = 121

a = 30

30 and 121 are your numbers

You have x+y = 115

and x = 19 + √y or x – 19 = √y.

Therefore x^2 – 38x + 281 = y – 115 – x

Solve the quadratic for x.

91