The two consecutive even integers are -8 and -6, or 6 and 8.
Introduction to Solving the Problem
This problem involves finding two consecutive even integers that satisfy a specific condition: the sum of their squares is 100. To solve this, we can use a systematic algebraic approach. The process involves setting up a quadratic equation and solving for the unknown variable.
Step 1: Defining the Integers and Forming the Equation
First, we define the two consecutive even integers using a variable. Let the first even integer be represented by xx
𝑥
. Since the integers are consecutive and even, the second even integer must be 2 greater than the first, so we can represent it as x+2x plus 2
𝑥+2
.
The problem states that the sum of the squares of these two integers is 100. We can translate this statement into the following algebraic equation:
x2+(x+2)2=100x squared plus open paren x plus 2 close paren squared equals 100
𝑥2+(𝑥+2)2=100
Step 2: Expanding and Simplifying the Equation
Next, we expand the squared term and combine like terms to transform the equation into a standard quadratic form, ax2+bx+c=0a x squared plus b x plus c equals 0
𝑎𝑥2+𝑏𝑥+𝑐=0
.
x2+(x2+4x+4)=100x squared plus open paren x squared plus 4 x plus 4 close paren equals 100
𝑥2+(𝑥2+4𝑥+4)=100
Combining the x2x squared
𝑥2
terms gives us:
2x2+4x+4=1002 x squared plus 4 x plus 4 equals 100
2𝑥2+4𝑥+4=100
To get the equation in standard form, we subtract 100 from both sides:
2x2+4x−96=02 x squared plus 4 x minus 96 equals 0
2𝑥2+4𝑥−96=0
To simplify the equation further, we can divide the entire equation by 2:
x2+2x−48=0x squared plus 2 x minus 48 equals 0
𝑥2+2𝑥−48=0
Step 3: Solving the Quadratic Equation
We now have a simplified quadratic equation, which can be solved by factoring. We are looking for two numbers that multiply to -48 and add to 2. These numbers are 8 and -6.
Factoring the quadratic equation gives us:
(x+8)(x−6)=0open paren x plus 8 close paren open paren x minus 6 close paren equals 0
(𝑥+8)(𝑥−6)=0
This equation holds true if either of the factors equals zero. This provides two possible solutions for xx
𝑥
.
-
Case 1: x+8=0⇒x=-8x plus 8 equals 0 implies x equals negative 8
𝑥+8=0⇒𝑥=−8
-
Case 2: x−6=0⇒x=6x minus 6 equals 0 implies x equals 6
𝑥−6=0⇒𝑥=6
Step 4: Finding the Pairs of Integers
We have two possible values for the first even integer, xx
𝑥
. We must now find the corresponding second even integer, x+2x plus 2
𝑥+2
, for each case.
-
**If x=6x equals 6
𝑥=6
:** The first integer is 6. The second consecutive even integer is 6+2=86 plus 2 equals 8
6+2=8
.
-
To verify: 62+82=36+64=1006 squared plus 8 squared equals 36 plus 64 equals 100
62+82=36+64=100
. This solution is correct.
-
-
**If x=-8x equals negative 8
𝑥=−8
:** The first integer is -8. The second consecutive even integer is -8+2=-6negative 8 plus 2 equals negative 6
−8+2=−6
.
-
To verify: (-8)2+(-6)2=64+36=100open paren negative 8 close paren squared plus open paren negative 6 close paren squared equals 64 plus 36 equals 100
(−8)2+(−6)2=64+36=100
. This solution is also correct.
-
Answer:
The two pairs of consecutive even integers whose squares sum to 100 are 6 and 8, and -8 and -6.
Show all