I Need Help Understanding The Type Of Problems That Go Like This; Find Two Numbers Whose Product Is 18 And Whose Sum Is 9?
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Usually, the answer to such a problem is a pair of integers. That means you can find the answer by examining the factors of the product value, 18 in this case.
18 = 1*18 = 2*9 = 3*6
You want the factors whose sum is 9.
1+18 = 19
2 + 9 = 11
3 + 6 = 9
The numbers you seek are 3 and 6.
The general way to solve a problem like this is to assign a variable to each number and write two equations. Let the numbers be represented by x and y.
x*y = 18 (two numbers have a product of 18)
x + y = 9 (the same two numbers have a sum of 9)
We can solve the second equation for y.
y = 9 - x (subtract x from both sides of the second equation above)
With this expression for y, we can make a substitution into the first equation.
x*(9 - x) = 18
This equation can be rearranged into the standard form for quadratics.
x*9 - x*x = 18 (use the distributive property)
9x = x^2 + 18 (add x^2 to both sides)
x^2 - 9x + 18 = 0 (subtract 9x from both sides, swap sides to put the 0 on the right)
This quadratic can be solved several ways. If you don't have a clue about the factors of 18, you can use the quadratic formula. Otherwise you can solve by factoring or by completing the square.
Factoring
(x - 3)(x - 6) = 0 (this method is essentially equivalent to the one we started with--finding factors of 18 that add to 9)
x = {3, 6}
Completing the square
x^2 - 9x + 18 + 2.25 = 2.25 (add a constant so that the constant term on the left is (9/2)^2)
(x - 4.5)^2 = (1.5)^2 (factor showing the squares)
x - 4.5 = ±1.5 (square root of both sides)
x = 4.5 ± 1.5 = {3, 6} (add 4.5 to both sides and evaluate)
Quadratic formula
For a*x^2 + b*x + c = 0, the solution is
x = (-b ±√(b^2 - 4*a*c))/(2*a)
= (-(-9) ±√((-9)^2 - 4*1*18))/(2*1)
= (9±√(81-72))/2
x = (9±3)/2 = {3, 6}
Each of the three methods above yields a value for x. You will note that if x is chosen as one of them, then y = 9-x will be the other one.
18 = 1*18 = 2*9 = 3*6
You want the factors whose sum is 9.
1+18 = 19
2 + 9 = 11
3 + 6 = 9
The numbers you seek are 3 and 6.
The general way to solve a problem like this is to assign a variable to each number and write two equations. Let the numbers be represented by x and y.
x*y = 18 (two numbers have a product of 18)
x + y = 9 (the same two numbers have a sum of 9)
We can solve the second equation for y.
y = 9 - x (subtract x from both sides of the second equation above)
With this expression for y, we can make a substitution into the first equation.
x*(9 - x) = 18
This equation can be rearranged into the standard form for quadratics.
x*9 - x*x = 18 (use the distributive property)
9x = x^2 + 18 (add x^2 to both sides)
x^2 - 9x + 18 = 0 (subtract 9x from both sides, swap sides to put the 0 on the right)
This quadratic can be solved several ways. If you don't have a clue about the factors of 18, you can use the quadratic formula. Otherwise you can solve by factoring or by completing the square.
Factoring
(x - 3)(x - 6) = 0 (this method is essentially equivalent to the one we started with--finding factors of 18 that add to 9)
x = {3, 6}
Completing the square
x^2 - 9x + 18 + 2.25 = 2.25 (add a constant so that the constant term on the left is (9/2)^2)
(x - 4.5)^2 = (1.5)^2 (factor showing the squares)
x - 4.5 = ±1.5 (square root of both sides)
x = 4.5 ± 1.5 = {3, 6} (add 4.5 to both sides and evaluate)
Quadratic formula
For a*x^2 + b*x + c = 0, the solution is
x = (-b ±√(b^2 - 4*a*c))/(2*a)
= (-(-9) ±√((-9)^2 - 4*1*18))/(2*1)
= (9±√(81-72))/2
x = (9±3)/2 = {3, 6}
Each of the three methods above yields a value for x. You will note that if x is chosen as one of them, then y = 9-x will be the other one.
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