Find two consecutive numbers if the square root of the smaller number is 3 less than the bigger number.
To find the two consecutive numbers, let's assign variables. Let's say the smaller number is x, and the bigger number is x + 1 (since they are consecutive).
According to the problem statement, the square root of the smaller number is 3 less than the bigger number. Mathematically, this can be expressed as:
√x = (x + 1) - 3
Now, let's solve the equation:
√x = x - 2
To eliminate the square root, we need to square both sides:
(√x)^2 = (x - 2)^2
x = (x - 2)(x - 2)
Expanding the right side:
x = x^2 - 4x + 4
Rearranging the equation to have all the terms on one side:
x - x^2 + 4x - 4 = 0
Combining like terms:
5x - x^2 - 4 = 0
To solve for x, we can either factor it or use the quadratic formula. In this case, factoring is simpler:
(x - 5)(x + 1) = 0
Setting each factor to zero:
x - 5 = 0 or x + 1 = 0
Solving for x in each case:
x = 5 or x = -1
Since we are looking for two consecutive numbers, we can discard the negative solution. Therefore, the smaller number is 5, and the bigger number is 5 + 1 = 6.
So, the two consecutive numbers are 5 and 6.
3 + sqrt x = x+1
sqrt x = x - 2
x = x^2 - 4 x + 4
x^2 - 5 x + 4 = 0
(x-4)(x-1) = 0
x = 1 or 4
check
if x = 4
2 = 5 - 3 yes so x = 4 and x+1 = 5
if x = 1, no way