The sum of the squares of three consecutive whole numbers is 77. Form a quadratic equation and hence solve it to determine the numbers. (Hint: let the three consecutive numbers be x, x+1 and x+2)
4, 5, 6
Let's solve the problem step-by-step:
Step 1: Formulating the equation:
From the given problem, we can assume the three consecutive whole numbers as x, x+1, and x+2.
According to the problem statement, the sum of the squares of these three numbers is 77.
Therefore, the equation can be formed as:
x^2 + (x+1)^2 + (x+2)^2 = 77
Step 2: Expanding the equation:
Simplifying the equation, we get:
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77
Combining like terms, the equation becomes:
3x^2 + 6x + 5 = 77
Step 3: Rearranging the equation:
Move 77 to the other side of the equation:
3x^2 + 6x + 5 - 77 = 0
Simplifying further, we have:
3x^2 + 6x - 72 = 0
Step 4: Solving the quadratic equation:
We can solve this quadratic equation by factoring or by using the quadratic formula.
Let's use the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 3, b = 6, and c = -72.
Substituting these values into the quadratic formula, we have:
x = (-6 ± √(6^2 - 4*3*(-72))) / (2*3)
x = (-6 ± √(36 + 864)) / 6
x = (-6 ± √900) / 6
x = (-6 ± 30) / 6
Now, we have two possible solutions:
For x = (-6 + 30) / 6 = 24 / 6 = 4,
and for x = (-6 - 30) / 6 = -36 / 6 = -6.
Step 5: Finding the consecutive numbers:
Now that we have obtained the values of x, we can find the three consecutive numbers.
For x = 4, the three consecutive numbers are 4, 5, and 6.
For x = -6, the three consecutive numbers are -6, -5, and -4.
Therefore, the two sets of numbers that satisfy the given condition are: 4, 5, 6 and -6, -5, -4.
To form a quadratic equation using the given problem, we need to use the hint and assume the three consecutive whole numbers as x, x+1, and x+2.
According to the problem statement, the sum of the squares of these three numbers is 77. We can express this mathematically as:
x^2 + (x+1)^2 + (x+2)^2 = 77
Expanding the equation:
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77
Combine like terms:
3x^2 + 6x + 5 = 77
Now we have the quadratic equation:
3x^2 + 6x + 5 - 77 = 0
Simplifying further:
3x^2 + 6x - 72 = 0
To solve this quadratic equation, we can factor it, complete the square, or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 3, b = 6, and c = -72.
Substituting these values into the quadratic formula:
x = (-6 ± √(6^2 - 4 * 3 * -72)) / (2 * 3)
Simplifying:
x = (-6 ± √(36 + 864)) / 6
x = (-6 ± √900) / 6
x = (-6 ± 30) / 6
Now, we have two possible solutions for x:
1) x = (-6 + 30) / 6 = 24 / 6 = 4
2) x = (-6 - 30) / 6 = -36 / 6 = -6
Therefore, the two possible sets of consecutive whole numbers are {4, 5, 6} and {-6, -5, -4}.
x^2 + (x+1)^2 + (x+2)^2=77
expand all of these, collect terms,
and solve.