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Chapter 1: Problem 74

Find all real solutions of the quadratic equation. $$0=x^{2}-4 x+1$$

Short Answer

Expert verified
The real solutions are \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \).

Step by step solution

01

Identify the quadratic equation

The given quadratic equation is \( x^2 - 4x + 1 = 0 \). We need to find the values of \( x \) for which this equation holds true.
02

Recall the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). In this equation, \( a = 1 \), \( b = -4 \), and \( c = 1 \).
03

Calculate the discriminant

The discriminant of the quadratic equation is given by \( \, D = b^2 - 4ac \). Substitute the values of \( a \), \( b \), and \( c \) into the discriminant formula: \( D = (-4)^2 - 4 \times 1 \times 1 = 16 - 4 = 12 \).
04

Solve for x using the quadratic formula

Now, substitute \( a = 1 \), \( b = -4 \), and \( D = 12 \) into the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{12}}{2 \times 1} = \frac{4 \pm \sqrt{12}}{2} \]Since \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \), the solutions become:\[ x = \frac{4 + 2\sqrt{3}}{2} \] and \[ x = \frac{4 - 2\sqrt{3}}{2} \]Simplifying these gives \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \).

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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 and an essential aspect of solving quadratic equations. These are equations of the form \( ax^2 + bx + c = 0 \). This formula allows us to find the roots or solutions of any quadratic equation, even when factoring is complicated or not possible. The quadratic formula is:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
How do we use it? First, identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. These coefficients are directly substituted into the formula:
  • The symbol \(\pm\) indicates that there will generally be two solutions, one using the plus and one using the minus.
  • The part \(b^2 - 4ac \) is known as the discriminant, which we'll cover later.
By remembering to correctly identify and substitute these parts, the quadratic formula provides a straightforward way to find solutions.
Discriminant
The discriminant is a crucial component within the quadratic formula, \( b^2 - 4ac \), and it helps determine the nature of the roots of the quadratic equation. The value of the discriminant tells us how many real solutions exist:
  • If \( D > 0 \), the quadratic equation has two distinct real roots.
  • If \( D = 0 \), there is exactly one real root, or a repeated root.
  • If \( D < 0 \), there are no real roots; however, there are two complex roots.
In our exercise, the discriminant was calculated as 12, which is greater than zero. This means that the quadratic equation \( x^2 - 4x + 1 = 0 \) has two distinct real solutions.
Real Solutions
Real solutions of a quadratic equation refer to the values of \( x \) that satisfy the equation where the discriminant, \( D \), is non-negative. Once the discriminant is determined, the real solutions are calculated using the quadratic formula:
  • The solutions are valid numbers on the real number line.
In the example provided, the roots calculated were: \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \). To simplify the solutions, one might initially write the expressions subtracting/adding the simplified \(\sqrt{12} \) as \( 2\sqrt{3} \), which results in clear, concise real numbers. Real solutions are especially significant because they can be graphed on the coordinate plane as x-intercepts of the quadratic function.

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Most popular questions from this chapter

\(x^{4}+a x^{2}+b\) A trinomial of the form \(x^{4}+a x^{2}+b\) can sometimes be factored easily. For example, $$ x^{4}+3 x^{2}-4=\left(x^{2}+4\right)\left(x^{2}-1\right) $$ But \(x^{4}+3 x^{2}+4\) cannot be factored in this way. Instead, we can use the following method. \(x^{4}+3 x^{2}+4=\left(x^{4}+4 x^{2}+4\right)-x^{2} \quad \begin{array}{l}\text { Add and } \\ \text { subtract } x^{2}\end{array}\) Factor perfect \(=\left(x^{2}+2\right)^{2}-x^{2}\) \(=\left[\left(x^{2}+2\right)-x\right]\left[\left(x^{2}+2\right)+x\right] \quad \begin{array}{l}\text { Difference of } \\ \text { squares }\end{array}\) \(=\left(x^{2}-x+2\right)\left(x^{2}+x+2\right)\) Factor the following, using whichever method is appropriate. (a) \(x^{4}+x^{2}-2\) (b) \(x^{4}+2 x^{2}+9\) (c) \(x^{4}+4 x^{2}+16\) (d) \(x^{4}+2 x^{2}+1\)

Multiply the algebraic expressions using a Special Product Formula and simplify. $$(r-2 s)^{2}$$

Factor the trinomial. $$x^{2}-6 x+5$$

Factor the expression completely. $$x^{2}+10 x+25$$

Perform the indicated operations and simplify. $$y^{1 / 3}\left(y^{2 / 3}+y^{5 / 3}\right)$$
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