Two numbers differ by 6. The sum of their squares is 116. Find the biggest number.
Let the bigger number b "y" and the other will be "y-6" as they differ by 6.
The sum of their squares=y^2+(y-6)^2=116
(y-6)^2=y^2 -2*6y + 6^2
So 2y^2 - 12y + 36 =116
Now solve for y.You can use quadratic formula.
x - y = 6.
X = y + 6
x^2 + y^2 = 116.
(y+6)^2 + y^2 = 116
y^2 + 12y + 36 + y^2 = 116
2y^2 + 12y + 36 = 116
Divide by 2:
y^2 + 6y + 18 = 58.
y^2 + 6y - 40 = 0.
(y-4)(y+10) = 0
y-4 = 0, Y = 4.
y+10 = 0, Y = -10.
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Use the positive value of Y: Y = 4.
x-y = 6
x - 4 = 6
X = 10.
Why did the number go on a diet? Because it wanted to be the biggest loser!
Let's solve this math problem and find the biggest number.
Let's assume that the two numbers are x and y, with x being the bigger number. We are given that the difference between these two numbers is 6, so we can write an equation:
x - y = 6
We are also given that the sum of their squares is 116, so we can write another equation:
x^2 + y^2 = 116
Now, we can use these two equations to solve for x and y.
From the first equation, we can solve for x:
x = y + 6
Substituting this expression for x into the second equation gives us:
(y + 6)^2 + y^2 = 116
Expanding and rearranging the equation, we have:
y^2 + 12y + 36 + y^2 = 116
Combining like terms:
2y^2 + 12y - 80 = 0
Dividing both sides by 2, we get:
y^2 + 6y - 40 = 0
Now we can solve this quadratic equation for y. Factoring it gives us:
(y - 4)(y + 10) = 0
Setting each factor equal to zero gives us two possible solutions for y:
y - 4 = 0, y = 4
y + 10 = 0, y = -10
Since y represents the smaller number, we ignore the negative solution. So, y = 4.
Substituting this value of y into the equation x = y + 6, we get:
x = 4 + 6
x = 10
So, the biggest number is x = 10.
Remember, math problems don't always have to be serious business!
Let's assume the two numbers are x and y. We are given two pieces of information:
1) Two numbers differ by 6: x - y = 6
2) The sum of their squares is 116: x^2 + y^2 = 116
To solve this system of equations, we can use the substitution method. First, let's solve the first equation for x:
x = y + 6
Now substitute this value of x into the second equation:
(y + 6)^2 + y^2 = 116
Expanding and simplifying the equation:
y^2 + 12y + 36 + y^2 = 116
2y^2 + 12y + 36 = 116
2y^2 + 12y - 80 = 0
Divide the entire equation by 2 to simplify it further:
y^2 + 6y - 40 = 0
Now, let's factor the quadratic equation:
(y + 10)(y - 4) = 0
Setting each factor equal to zero and solving for y:
y + 10 = 0 OR y - 4 = 0
y = -10 OR y = 4
Since we are looking for the biggest number, we take y = 4. Substituting this value back into the equation x = y + 6, we find:
x = 4 + 6 = 10
So, the biggest number is 10.
To solve this problem, we can start by assigning variables to the two numbers. Let's say the bigger number is x and the smaller number is y. Given that the numbers differ by 6, we can express this as an equation:
x - y = 6 (Equation 1)
We're also told that the sum of their squares is 116, so we can write another equation:
x^2 + y^2 = 116 (Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) with two variables (x and y). We can solve this system to find the values of x and y, and then determine which one is the bigger number.
Let's solve Equation 1 for x by adding y to both sides:
x = y + 6
Now we can substitute this expression for x in Equation 2:
(y + 6)^2 + y^2 = 116
Expanding and simplifying:
y^2 + 12y + 36 + y^2 = 116
2y^2 + 12y + 36 = 116
Rearranging the equation by subtracting 116 from both sides:
2y^2 + 12y + 36 - 116 = 0
2y^2 + 12y - 80 = 0
Dividing the entire equation by 2:
y^2 + 6y - 40 = 0
Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, we can factor the equation:
(y + 10)(y - 4) = 0
This gives us two possible solutions for y:
y + 10 = 0 or y - 4 = 0
y = -10 or y = 4
Now we can substitute these values into Equation 1 to find the corresponding values for x:
x - (-10) = 6 or x - 4 = 6
x + 10 = 6 or x = 10
x = -4 or x = 10
We have two pairs of solutions, (-4, -10) and (10, 4).
From these solutions, we can see that the biggest number is 10.