Problem 41 Solve. $$4 x^{2}=4 x+1$$... [FREE SOLUTION]

91影视

Calculus and Its Applications

Marvin L. Bittinger, David J. Ellenbogen, Scott A. Surgent

$Math Studyset 91影视 Explanations$ Math

10 Edition

Chapter 0: Problem 41

Solve. $$4 x^{2}=4 x+1$$

Short Answer

Expert verified

The solutions are $ x = \frac{1 + \sqrt{2}}{2} $ and $ x = \frac{1 - \sqrt{2}}{2} $.

Step by step solution

Rewrite the Equation in Standard Form

Subtract $4x + 1$ from both sides of the equation to set it equal to zero: \[4x^2 - 4x - 1 = 0\]

Identify Coefficients for Quadratic Formula

For the quadratic equation $ax^2 + bx + c = 0$, identify the coefficients: $a = 4$, $b = -4$, and $c = -1$.

Apply the Quadratic Formula

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ by plugging in the coefficients: \[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-1)}}{2(4)}\]

Simplify Under the Square Root

Calculate the expression under the square root: \[x = \frac{4 \pm \sqrt{16 + 16}}{8}\]

Simplify the Square Root

Simplify the expression under the square root: \[x = \frac{4 \pm \sqrt{32}}{8}\]

Simplify the Expression Under the Square Root Further

Since $ \sqrt{32} = 4\sqrt{2} $, \[x = \frac{4 \pm 4\sqrt{2}}{8}\]

Simplify the Entire Expression

Divide each term in the numerator by 8: \[x = \frac{4}{8} \pm \frac{4\sqrt{2}}{8} = \frac{1}{2} \pm \frac{\sqrt{2}}{2}\]

State the Final Solutions

The solutions to the equation are: \[x = \frac{1 + \sqrt{2}}{2}\] and \[x = \frac{1 - \sqrt{2}}{2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form $ax^2 + bx + c = 0$. The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula helps us find the values of $x$ that satisfy the equation. Let's break it down:

$a, b,$ and $c$ are the coefficients of the equation.
$b^2 - 4ac$ is called the discriminant. It determines the nature of the roots.

By plugging in the coefficients into this formula, we can solve any quadratic equation. It makes the process straightforward, even if the quadratic doesn't factor nicely.
We saw this in the original solution where we used the quadratic formula to solve $4x^2 - 4x - 1 = 0$.

standard form

In order to use the quadratic formula, the quadratic equation must be in its standard form:
$ax^2 + bx + c = 0$
This is crucial because it allows us to identify the coefficients $a, b,$ and $c$. Let's look at the given equation:
$4x^2 = 4x + 1$
Step 1 involves converting this into standard form by moving all terms to one side:
$4x^2 - 4x - 1 = 0$
Now we clearly see that:

$a = 4$
$b = -4$
$c = -1$

This setup is essential for correctly applying the quadratic formula. Without this form, correctly identifying coefficients would be difficult.

solving quadratic equations

Solving quadratic equations means finding values of $x$ that make the equation true. Here are the general approaches:

**Factoring**: Useful if the equation can be factored easily.
**Quadratic Formula**: Works for all forms, especially when factoring is complex.
**Completing the Square**: Another method that rewrites the equation in a way that makes it easier to solve.

Consider our given equation:
$4x^2 - 4x - 1 = 0$
Through our solution steps, we applied the quadratic formula because the equation didn't factor easily. We identified coefficients, substituted them into the formula, and simplified step-by-step to reach our solutions.

coefficients

Coefficients are the numerical factors of the terms in a polynomial. For a quadratic equation $ax^2 + bx + c = 0$, they are:

$a$ is the coefficient of $x^2$.
$b$ is the coefficient of $x$.
$c$ is the constant term (coefficient of $x^0$).

Identifying and correctly substituting these is key to solving quadratics using the quadratic formula. In our problem, we had the equation transformed to:
$4x^2 - 4x - 1 = 0$
Here,

$a = 4$
$b = -4$
$c = -1$

True understanding of coefficients helps in accurate calculation and successful solution of the quadratic equation.

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