Solving a Quadratic Equation: A Sum of a Number and Twice Its Square Equals 3

Let's explore a classic algebraic problem involving a quadratic equation. We will solve the equation where a number, ( x ), plus twice its square equals 3. This sum can be expressed as ( x 2x^2 3 ), or rearranged to a standard quadratic form: ( 2x^2 x - 3 0 ). By solving this equation, we can determine the values of ( x ).

Problem Statement

Given the equation: 2x^2 x - 3 0

Solution Using the Quadratic Formula

The quadratic formula is a powerful tool for solving such equations. The quadratic formula is:

x frac{-b pm sqrt{b^2 - 4ac}}{2a}

In our equation, we identify the coefficients as follows:

Plugging these values into the quadratic formula, we get:

x frac{-1 pm sqrt{1^2 - 4 cdot 2 cdot (-3)}}{2 cdot 2}

First, calculate the discriminant:

1^2 - 4 cdot 2 cdot (-3) 1 24 25

Next, substitute the discriminant back into the quadratic formula:

x frac{-1 pm sqrt{25}}{4}

Since the square root of 25 is 5, we have two solutions:

x frac{-1 5}{4} frac{4}{4} 1

x frac{-1 - 5}{4} frac{-6}{4} -frac{3}{2}

Thus, the solutions are:

x 1 quad text{and} quad x -frac{3}{2}

Validation

To validate the solutions, we substitute them back into the original equation ( x 2x^2 3 ).

For ( x 1 )

( 1 2(1)^2 1 2 3 )

For ( x -frac{3}{2} )

( -frac{3}{2} 2left(-frac{3}{2}right)^2 -frac{3}{2} 2left(frac{9}{4}right) -frac{3}{2} frac{18}{4} -frac{3}{2} frac{9}{2} frac{6}{2} 3 )

Both solutions satisfy the original equation, confirming our results.

Conclusion

The number that satisfies the equation ( x 2x^2 3 ) is:

boxed{1} quad text{or} quad boxed{-frac{3}{2}}

Understanding and solving quadratic equations is crucial for many areas of mathematics and real-world applications. By applying the quadratic formula, we can find accurate and reliable solutions to such problems.