the tens digit of a certain number is 4 more than the units digit. the sum of the squares of the two digits is 26. find the number.
Well, well, well, let's solve this riddle, shall we? So, we have a number where the tens digit is 4 more than the units digit. To ease things up, let's represent the tens digit as "X" and the units digit as "Y." According to our riddle, we have the equation X = Y + 4.
Now, the second clue tells us that the sum of the squares of the two digits is 26. So, we have another equation: X^2 + Y^2 = 26.
Now, let's substitute the value we found for X (which is Y + 4) into our second equation:
(Y + 4)^2 + Y^2 = 26.
Expanding and simplifying the equation, we'll get:
Y^2 + 8Y + 16 + Y^2 = 26.
Combining like terms, the equation becomes:
2Y^2 + 8Y + 16 = 26.
Now, let's subtract 26 from both sides:
2Y^2 + 8Y - 10 = 0.
Divide everything by 2:
Y^2 + 4Y - 5 = 0.
Now, let's factor this quadratic equation:
(Y + 5)(Y - 1) = 0.
Solving for Y, we have two possible solutions: Y = -5 or Y = 1.
Since we're dealing with digits, we can discard the negative solution. So, Y = 1.
Now, going back to our equation X = Y + 4, we can substitute Y = 1:
X = 1 + 4 = 5.
Voila! The number we're looking for is 51. Ta-da!
Hope my mathematical clownery didn't confuse you too much!
Let's call the tens digit "x" and the units digit "y".
According to the problem, x = y + 4, and the sum of the squares of the two digits is 26, so x^2 + y^2 = 26.
We can substitute the value of x from the first equation into the second equation:
(y + 4)^2 + y^2 = 26
Expanding the equation:
y^2 + 8y + 16 + y^2 = 26
Combining like terms:
2y^2 + 8y + 16 = 26
Rearranging the equation:
2y^2 + 8y + 16 - 26 = 0
Simplifying:
2y^2 + 8y - 10 = 0
Dividing the equation by 2 to simplify:
y^2 + 4y - 5 = 0
Using the quadratic formula to solve for y:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 4, and c = -5. Substituting the values:
y = (-4 ± √(4^2 - 4(1)(-5))) / (2(1))
Simplifying inside the square root:
y = (-4 ± √(16 + 20)) / 2
y = (-4 ± √36) / 2
y = (-4 ± 6) / 2
Now we have two possible solutions for y:
1. y = (-4 + 6) / 2 = 2 / 2 = 1
2. y = (-4 - 6) / 2 = -10 / 2 = -5
Since the units digit cannot be negative, we can disregard the second solution.
Now we can substitute the value of y into the equation x = y + 4 to find x:
x = 1 + 4 = 5
Therefore, the number is 51.
To find the number, let's assign variables to the tens and units digits.
Let the tens digit be represented by the variable 't', and the units digit be represented by the variable 'u'.
We are given two conditions:
1. The tens digit is 4 more than the units digit: t = u + 4.
2. The sum of the squares of the digits is 26: t^2 + u^2 = 26.
Now, let's solve the system of equations using substitution.
Substitute the value of 't' from the first equation into the second equation:
(u + 4)^2 + u^2 = 26
Expand:
u^2 + 8u + 16 + u^2 = 26
Combine like terms:
2u^2 + 8u + 16 = 26
Rearrange the equation:
2u^2 + 8u + 16 - 26 = 0
2u^2 + 8u - 10 = 0
Divide by 2 to simplify:
u^2 + 4u - 5 = 0
Now, we can factor this quadratic equation:
(u + 5)(u - 1) = 0
Solve for 'u':
u + 5 = 0 or u - 1 = 0
u = -5 or u = 1
Since 'u' represents a digit, it cannot be negative. Therefore, 'u' is 1.
Substitute the value of 'u' back into the first equation to find 't':
t = u + 4
t = 1 + 4
t = 5
So, the tens digit is 5 and the units digit is 1. Therefore, the number is 51.
unit digit --- x
tens digit ----x+4
x^2 + (x+4)^2 = 26^2
solve the quadratic