Problem 21
Question
In \(15-23 :\) a. Find the value of the discriminant and determine if the roots of the quadratic equation are \((1)\) rational and unequal, \((2)\) rational and equal, \((3)\) irrational and unequal, or \((4)\) not real numbers. b. Use any method to find the real roots of the equation if they exist. $$ 4 x-1=4 x^{2} $$
Step-by-Step Solution
Verified Answer
The roots are real, rational, and equal: \(x = \frac{1}{2}\).
1Step 1: Write the Quadratic Equation in Standard Form
First, we need to rearrange the equation \(4x - 1 = 4x^2\) into the standard quadratic form, \(ax^2 + bx + c = 0\). This involves moving all terms to one side:\[4x^2 - 4x + 1 = 0\]
2Step 2: Calculate the Discriminant
The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). For our equation, \(a = 4\), \(b = -4\), and \(c = 1\). Substitute these values into the formula:\[D = (-4)^2 - 4 \cdot 4 \cdot 1 = 16 - 16 = 0\]
3Step 3: Interpret the Discriminant
The value of the discriminant is \(0\). A discriminant of \(0\) indicates that the quadratic equation has real, rational, and equal roots. In this case, it falls under category (2): rational and equal roots.
4Step 4: Find the Roots of the Equation
Since the discriminant is \(0\), the roots can be found using the formula \(x = \frac{-b \pm \sqrt{D}}{2a}\). We know \(D = 0\), so the formula simplifies to:\[x = \frac{-b}{2a}\]Substituting our values \(b = -4\) and \(a = 4\):\[x = \frac{-(-4)}{2 \cdot 4} = \frac{4}{8} = \frac{1}{2}\]This means the equation has one real, rational, and equal root: \(x = \frac{1}{2}\).
Key Concepts
Quadratic EquationReal RootsRational RootsStandard Form
Quadratic Equation
A quadratic equation is a type of polynomial equation that you often encounter in algebra. Its general form is given by: \[ ax^2 + bx + c = 0 \] Here,
- \(a\), \(b\), and \(c\) are constants,
- \(a eq 0\) because if \(a = 0\), the equation would not be quadratic,
- \(x\) represents the variable or unknown we want to solve for.
Real Roots
Real roots of a quadratic equation are the values of \(x\) that satisfy the equation by making it equal to zero. These roots can be found using several methods:
- Factoring: Express the quadratic as a product of two binomials, if possible.
- Quadratic Formula: Use the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Completing the square: Rewrite the equation to make it a perfect square trinomial.
Rational Roots
Rational roots are specific real roots that can be expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers, and \(q eq 0\). Rational roots occur depending on the discriminant's value:
- If \(D = 0\), there is exactly one rational and repeated root given by \(x = \frac{-b}{2a}\).
- If \(D > 0\) and is a perfect square, the roots are rational and unequal.
- If \(D < 0\), there are no rational roots.
Standard Form
The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). This form is essential for a structured approach to solving and analyzing quadratic equations.Here's why the standard form is important:
- Identify Coefficients: It makes it easy to identify \(a\), \(b\), and \(c\) for use in the quadratic formula.
- Discriminant Calculation: It aligns with the discriminant formula \(D = b^2 - 4ac\), necessary for determining the nature of roots.
- Simplification: Equation manipulation tasks become systematic, aiding in employing consistent strategies like factoring or completing the square.
Other exercises in this chapter
Problem 21
In \(18-35,\) find each common solution algebraically. Express irrational roots in simplest radical form. $$ \begin{array}{l}{y=x^{2}-8 x+6} \\ {2 x-y=10}\end{a
View solution Problem 21
In \(19-28 :\) a. Find \(\mathrm{f}(a)\) for each given function. b. Is \(a\) a root of the function? $$ \mathrm{f}(x)=5 x^{2}+4 x+1 \text { and } a=-1 $$
View solution Problem 21
In \(18-25,\) write the complex conjugate of each number. $$ -6-9 i $$
View solution Problem 21
In \(19-34,\) write each sum or difference in terms of \(i\) $$ \sqrt{-144}+\sqrt{-1} $$
View solution