The sum of the squares of three consecutive integers is 194. what are the integers?
Fist number= a
Second number= a+1
Third number= a+2
a^2+(a+1)^2+(a+2)^2=194
(a+1)^2=a^2+2*a*1+1^2=a^2+2a+1
(a+2)^2=a^2*2a*2+2^2=a^2+4a+4
a^2+a^2+2a+1+a^2+4a+4=194
3a^2+6a+1+4=194
3a^2+6a+5-194=0
3a^2+6a-189=0
The exactsolutions of this equation are:
a=7 and a=-9
Solution:
First number 7
Second number 7+1=8
Third number 7+2=9
7^2+8^2+9^2=49+64+81=194
how did u get a to b 7 and 9
In google type:
"quadratic equation online"
When you see list of result click on:
webgraphingcom/quadraticequation_quadraticformula.jsp
When this page be open in rectacangle type equation:
3a^2+6a-189=0
and click option solve it!
You will see solutions of this equation Step-by-step
To solve this problem, let's use algebraic equations.
Let's assume the three consecutive integers are x, x+1, and x+2, respectively.
The sum of their squares is:
x^2 + (x+1)^2 + (x+2)^2 = 194
Expanding the equation:
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 194
Combining like terms:
3x^2 + 6x + 5 = 194
Rearranging the equation:
3x^2 + 6x - 189 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use factoring.
Factoring the quadratic equation:
3(x^2 + 2x - 63) = 0
(x^2 + 2x - 63) = 0
Now, we can factor the quadratic expression:
(x - 7)(x + 9) = 0
Setting each factor equal to zero:
x - 7 = 0 or x + 9 = 0
Solving for x:
x = 7 or x = -9
Therefore, the two possible sets of three consecutive integers whose sum of squares is 194 are:
1. 7, 8, 9
2. -9, -8, -7