Twice the square of a positive real number is equal to 4 more than the number. What is the number?
2 x^2 = 4 + x
2 x^2 - x - 4 = 0
about 1.69
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Let's solve this problem step by step.
Let's assume the positive real number as "x".
According to the given condition, twice the square of the number is equal to 4 more than the number.
Mathematically, this can be written as:
2x^2 = x + 4
Next, let's solve this equation.
2x^2 - x - 4 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.
By factoring, the equation can be written as:
(2x + 4)(x - 1) = 0
From this equation, we get two solutions:
1) 2x + 4 = 0
=> 2x = -4
=> x = -2
2) x - 1 = 0
=> x = 1
Since we are looking for a positive real number, we can discard the solution x = -2.
Therefore, the positive real number satisfying the given condition is x = 1.
To solve this problem, let's start by assigning a variable to the unknown number. Let's call the number "x".
The problem states that twice the square of a positive real number (2x²) is equal to 4 more than the number (x + 4):
2x² = x + 4
Now, we can rearrange the equation to solve for x:
2x² - x - 4 = 0
This is a quadratic equation in the form of ax² + bx + c = 0, where a = 2, b = -1, and c = -4.
To solve this quadratic equation, we can factorize or use the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax² + bx + c = 0, the solution for x is given by:
x = (-b ± √(b² - 4ac)) / (2a)
Substituting the values from our equation into the quadratic formula, we have:
x = (-(-1) ± √((-1)² - 4 * 2 * (-4))) / (2 * 2)
x = (1 ± √(1 + 32)) / 4
x = (1 ± √33) / 4
Therefore, the two possible solutions for x are:
x₁ = (1 + √33) / 4
x₂ = (1 - √33) / 4
Since we are given that x is a positive real number, we can disregard the negative solution x₂. Thus, the number is:
x = (1 + √33) / 4