Find a positive real number such that its square is equal to 4 times the number, increased by 60.
Did you translate the English into Math?
I said:
x^2 = 4x + 60
now you have a nice and simple quadratic which factors.
x^2 - 4x - 60 = 0
solve for x, and take only the positive result
Let's represent the positive real number by "x". According to the given information, we can write the equation as x^2 = 4x + 60.
To solve this equation, we'll set it equal to zero by subtracting 4x and 60 from both sides:
x^2 - 4x - 60 = 0
Now, we'll solve this quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
In our equation, a = 1, b = -4, and c = -60.
x = (-(-4) ± √((-4)^2 - 4(1)(-60))) / (2(1))
= (4 ± √(16 + 240)) / 2
= (4 ± √256) / 2
= (4 ± 16) / 2
Now, we have two possible solutions for x:
1) When we use the plus sign in ±: (4 + 16) / 2 = 20 / 2 = 10
2) When we use the minus sign in ±: (4 - 16) / 2 = -12 / 2 = -6
Since we are looking for a positive real number, the solution is x = 10.
To find a positive real number that satisfies the given condition, let's use algebraic notation.
Let's assume the positive real number we are looking for is denoted by "x". According to the given condition, its square is equal to 4 times the number, increased by 60.
Mathematically, we can represent this as:
x^2 = 4x + 60
To solve this equation, we need to bring all terms to one side and set the equation equal to zero:
x^2 - 4x - 60 = 0
Now we can solve this quadratic equation by factoring or applying the quadratic formula.
Factoring:
The quadratic equation x^2 - 4x - 60 = 0 can be factored as (x - 10)(x + 6) = 0.
Setting each factor equal to zero, we have two possible solutions:
x - 10 = 0, which gives x = 10
x + 6 = 0, which gives x = -6
Since we are looking for a positive real number, x = -6 is not a valid solution. Therefore, the positive real number that satisfies the given condition is x = 10.
Thus, the positive real number we are searching for is 10.