Problem 3
Question
Use the discriminant to determine the number of real solutions of the quadratic equation. \(3 x^{2}+4 x+1=0\)
Step-by-Step Solution
Verified Answer
The given quadratic equation has two real solutions.
1Step 1: Identify the coefficients
In the given quadratic equation \(3x^2 + 4x + 1 = 0\), identify the values of a, b and c. Here, \(a = 3\), \(b = 4\), and \(c = 1\).
2Step 2: Calculate the discriminant
The discriminant is calculated by the formula \(b^2 - 4ac\). Substituting the given values, we get \((4)^2 - 4*3*1 = 16 - 12 = 4\).
3Step 3: Determine the number of real solutions
As the discriminant is greater than zero (4 > 0), the given quadratic equation will have 2 real solutions.
Key Concepts
Discriminant and Real SolutionsSolving Quadratic EquationsQuadratic Formula
Discriminant and Real Solutions
The discriminant is a crucial element in understanding the nature of the solutions to a quadratic equation, such as the one in our exercise: \(3x^2 + 4x + 1 = 0\). Simplified, the discriminant is a part of a formula that provides insight into the number and type of solutions for a quadratic equation. To calculate it, we use the coefficient values of a quadratic equation in the form \(ax^2 + bx + c = 0\), and insert them into the discriminant formula \(D = b^2 - 4ac\).
The value of the discriminant reveals:
The value of the discriminant reveals:
- If \(D > 0\), there are two distinct real solutions.
- If \(D = 0\), there is exactly one real solution (real and repeated).
- If \(D < 0\), there are no real solutions; instead, there are two complex solutions.
Solving Quadratic Equations
Quadratic equations are a foundational concept in algebra, and solving them is a fundamental skill for students. A quadratic equation takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). There are several methods for solving quadratic equations including factoring, using square roots, completing the square, and employing the quadratic formula.
Factoring involves writing the equation as a product of its factors. The square root method can be used when the quadratic equation is a perfect square. Completing the square transforms the equation into a perfect square trinomial, allowing for the use of square roots to solve. However, these methods are not always applicable or straightforward, which brings us to a universal method: the quadratic formula.
Factoring involves writing the equation as a product of its factors. The square root method can be used when the quadratic equation is a perfect square. Completing the square transforms the equation into a perfect square trinomial, allowing for the use of square roots to solve. However, these methods are not always applicable or straightforward, which brings us to a universal method: the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula quickly provides the solutions by incorporating our previously calculated discriminant \((b^2 - 4ac)\) under the square root symbol. The symbol \(\pm\) indicates that there are two solutions: one from adding the square root value to \(-b\), and one from subtracting it. Applied to our example, our solutions using the quadratic formula are:
\[ x = \frac{-4 \pm \sqrt{4}}{2 \cdot 3} \]
By simplifying, we find two real solutions for \(x\). This formula ensures that we can find solutions to any quadratic equation, whether they are real or complex, without the need for laborious factoring or other methods. It's a comprehensive solution to a common algebraic challenge.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula quickly provides the solutions by incorporating our previously calculated discriminant \((b^2 - 4ac)\) under the square root symbol. The symbol \(\pm\) indicates that there are two solutions: one from adding the square root value to \(-b\), and one from subtracting it. Applied to our example, our solutions using the quadratic formula are:
\[ x = \frac{-4 \pm \sqrt{4}}{2 \cdot 3} \]
By simplifying, we find two real solutions for \(x\). This formula ensures that we can find solutions to any quadratic equation, whether they are real or complex, without the need for laborious factoring or other methods. It's a comprehensive solution to a common algebraic challenge.
Other exercises in this chapter
Problem 3
Write an inequality that represents the interval. Then state whether the interval is bounded or unbounded. \((11, \infty)\)
View solution Problem 3
Find the real solution(s) of the polynomial equation. Check your solutions. \(4 x^{4}-18 x^{2}=0\)
View solution Problem 3
Write the quadratic equation in general form. $$ x^{2}=25 x $$
View solution Problem 3
Write an algebraic expression for the verbal expression. The distance traveled in \(t\) hours by a car traveling at 50 miles per hour
View solution