A Quadratic Word Problem.
The difference between two positive numbers is 3. The sum of the squares of the numbers is 89. Find the numbers.
a = first number
b = second number
a - b = 3 Add b to both sides
a - b + b = 3 + b
a = b + 3
a ^ 2 + b ^ 2 = 89
( b + 3 ) ^ 2 + b ^ 2 = 89
b ^ 2 + 2 * b * 3 + 3 ^ 2 + b ^ 2 = 89
b ^ 2 + 6 b + 9 + b ^ 2 = 89
2 b ^ 2 + 6 b + 9 = 89 Divide both sides by 2
b ^ 2 + 3 b + 9 / 2 = 89 / 2 Subtract 9 / 2 to both sides
b ^ 2 + 3 b + 9 / 2 - 9 / 2 = 89 / 2 - 9 / 2
b ^ 2 + 3 b = 80 / 2
b ^ 2 + 3 b = 40 Add 9 / 4 to both sides
b ^ 2 + 3 b + 9 / 4 = 40 + 9 / 4
b ^ 2 + 3 b + 9 / 4 = 160 / 4 + 9 / 4
b ^ 2 + 3 b + 9 / 4 = 169 / 4
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Remark :
b ^ 2 + 3 b + 9 / 4 = ( b + 3 / 2 ) ^ 2
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( b + 3 / 2 ) ^ 2 = 169 / 4 take sqrt to both sides
b + 3 / 2 = 13 / 2 Subtract 3 / 2 to both sides
b + 3 / 2 - 3 / 2 = 13 / 2 - 3 / 2
b = 10 / 2 = 5
a = b + 3 = 5 + 3 = 8
Nuumbers are :
5 and 8
Proof :
8 - 5 = 3
8 ^ 2 + 5 ^ 2 = 64 + 25 = 89
The product of two number is between 40 and 50. One number is 4 time less than the other
To solve this quadratic word problem, we can set up a system of equations.
Let's assume the two positive numbers as x and y.
According to the problem, the difference between the two positive numbers is 3. So, we can write the equation as:
x - y = 3 ----(Equation 1)
The sum of the squares of the numbers is 89. So, we can write the equation as:
x^2 + y^2 = 89 ----(Equation 2)
We now have a system of two equations (Equation 1 and Equation 2) which we can solve simultaneously to find the values of x and y.
Method 1: Substitution Method
We can solve Equation 1 for x and substitute it into Equation 2:
x = y + 3
(y + 3)^2 + y^2 = 89
Expanding and simplifying:
y^2 + 6y + 9 + y^2 = 89
2y^2 + 6y + 9 = 89
2y^2 + 6y - 80 = 0
Now, we have a quadratic equation in standard form. To solve it, we can factor or use the quadratic formula. Factoring might be a bit challenging for this particular equation, so let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = 6, and c = -80.
y = (-6 ± √(6^2 - 4(2)(-80))) / (2(2))
Solving the quadratic formula gives us two values of y:
y1 = (-6 + √496) / 4 ≈ 6.24
y2 = (-6 - √496) / 4 ≈ -10.74 (extraneous solution since it's given that the numbers are positive)
Now we can substitute the value of y back into Equation 1 to find x:
x - (6.24) = 3
x ≈ 9.24
Thus, the two positive numbers are approximately 6.24 and 9.24.
Method 2: Elimination Method
We can eliminate one variable by squaring both sides of Equation 1:
(x - y)^2 = (3)^2
x^2 - 2xy + y^2 = 9 ----(Equation 3)
Subtracting Equation 3 from Equation 2, we get:
(x^2 + y^2) - (x^2 - 2xy + y^2) = 89 - 9
4xy = 80
xy = 20 ----(Equation 4)
Now, we have two equations (Equation 1 and Equation 4) with two variables, x and y. We can solve this system of equations.
From Equation 4, we have xy = 20. Substituting this value in Equation 1, we get:
x - y = 3
x = y + 3
Now we can substitute the value of x in terms of y in Equation 4:
(y + 3)y = 20
y^2 + 3y - 20 = 0
Now, we have another quadratic equation in standard form. We can factor or use the quadratic formula to solve it:
(y + 5)(y - 4) = 0
y = -5 or y = 4
Since it is given that the numbers are positive, we can ignore the negative solution, y = -5.
Now we can substitute the value of y in Equation 1 to find x:
x - (4) = 3
x = 7
Thus, the two positive numbers are 4 and 7.
Therefore, the two solutions to the problem are approximately 6.24 and 9.24 OR 4 and 7.