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