Problem 23
Question
Factor to find the \(x\)-intercepts of the parabola described by the quadratic function. Also find the real zeros of the function. $$g(x)=2 x^{2}+5 x-3$$
Step-by-Step Solution
Verified Answer
The x-intercepts of the parabola described by the quadratic function \(g(x)=2 x^{2}+5 x-3\) are (-3,0) and (0.5,0). The zeros of the function are -3 and 0.5.
1Step 1: Identify the coefficients
Identify the coefficients a, b, and c in the quadratic equation \(g(x)=2 x^{2}+5 x-3\). Here, a = 2, b = 5, and c = -3.
2Step 2: Apply the quadratic formula
Apply the quadratic formula, which is \(x = \frac{-b\pm \sqrt{b^{2} - 4ac}}{2a}\). Substituting the values of a, b, and c, get \(x = \frac{-5\pm \sqrt{5^{2} - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}\). This simplifies to \(x = \frac{-5\pm \sqrt{25 + 24}}{4}\).
3Step 3: Solve for x
Solving for x, get \(x = \frac{-5\pm \sqrt{49}}{4}\). This simplifies to \(x = \frac{-5\pm 7}{4}\). Therefore, the solutions are \(x = \frac{-5-7}{4}\) and \(x = \frac{-5+7}{4}\). Simplifying these, find \(x = -3\) and \(x = 0.5\). So, the x-intercepts are (-3,0) and (0.5,0), and the real zeros of the function are -3 and 0.5.
Key Concepts
Factoring QuadraticsQuadratic FormulaReal Zeros of Functions
Factoring Quadratics
Factoring quadratics is a method used to express a quadratic equation in the form of two binomials multiplied together. Consider the quadratic function given by the exercise: \(g(x) = 2x^{2} + 5x - 3\). The idea is to find two numbers that multiply to the product of the coefficient of \(x^2\) (a) and the constant term (c), while simultaneously adding up to the coefficient of \(x\) (b).
\(g(x) = 2x^{2} + 6x - 1x - 3\).
Then, group the terms to factor by grouping:
\(g(x) = (2x^{2} + 6x) + (-1x - 3)\).
Factor each group:
\(g(x) = 2x(x + 3) - 1(x + 3)\).
Finally, factor out the common binomial:
\(g(x) = (2x - 1)(x + 3)\).
This tells us that the factors of the quadratic are \((2x - 1)\) and \((x + 3)\).
- In our equation, \(a = 2\), \(b = 5\), and \(c = -3\).
- We look for two numbers that multiply to \(2 \, * \, (-3) = -6\) and add to 5.
- The numbers that work are 6 and -1 because \(6 * (-1) = -6\) and \(6 + (-1) = 5\).
\(g(x) = 2x^{2} + 6x - 1x - 3\).
Then, group the terms to factor by grouping:
\(g(x) = (2x^{2} + 6x) + (-1x - 3)\).
Factor each group:
\(g(x) = 2x(x + 3) - 1(x + 3)\).
Finally, factor out the common binomial:
\(g(x) = (2x - 1)(x + 3)\).
This tells us that the factors of the quadratic are \((2x - 1)\) and \((x + 3)\).
Quadratic Formula
The quadratic formula is a universal tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It states that the solutions for \(x\) can be found using:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula is particularly helpful when the quadratic doesn't easily factor, or when factoring proves to be challenging.
Let's apply this formula to our quadratic \(g(x) = 2x^{2} + 5x - 3\):
- Here, \(a = 2\), \(b = 5\), and \(c = -3\).
- Substitute these values into the quadratic formula:
\[x = \frac{-5 \pm \sqrt{5^2 - 4 * 2 * (-3)}}{2 * 2}\]
- Simplify inside the square root:
\(5^2 = 25\) and \(-4 * 2 * (-3) = 24\), so the expression becomes \(25 + 24 = 49\).
- Applying the square root, \(\sqrt{49} = 7\).
- Substitute back, and simplify:
\[x = \frac{-5 \pm 7}{4}\]
This results in two solutions:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula is particularly helpful when the quadratic doesn't easily factor, or when factoring proves to be challenging.
Let's apply this formula to our quadratic \(g(x) = 2x^{2} + 5x - 3\):
- Here, \(a = 2\), \(b = 5\), and \(c = -3\).
- Substitute these values into the quadratic formula:
\[x = \frac{-5 \pm \sqrt{5^2 - 4 * 2 * (-3)}}{2 * 2}\]
- Simplify inside the square root:
\(5^2 = 25\) and \(-4 * 2 * (-3) = 24\), so the expression becomes \(25 + 24 = 49\).
- Applying the square root, \(\sqrt{49} = 7\).
- Substitute back, and simplify:
\[x = \frac{-5 \pm 7}{4}\]
This results in two solutions:
- \(x = \frac{-5 + 7}{4} = 0.5\)
- \(x = \frac{-5 - 7}{4} = -3\)
Real Zeros of Functions
The real zeros of a function are the values of \(x\) where the function intersects the x-axis. In simple terms, they are the x-values where \(g(x) = 0\). Finding these zeros can help us understand the behavior and shape of the parabola described by a quadratic function.
The quadratic function \(g(x) = 2x^{2} + 5x - 3\) has its real zeros or x-intercepts where we set \(g(x) = 0\). These are the points that satisfy the equation \((2x - 1)(x + 3) = 0\) derived from our factoring.
- For \((2x - 1) = 0\), solving gives \(x = 0.5\).
- For \((x + 3) = 0\), solving gives \(x = -3\).
Thus, the real zeros are \(x = -3\) and \(x = 0.5\). These are also the x-values that you would see where the parabola crosses the x-axis on a graph. Real zeros provide insights into the nature of quadratic expressions and their graphical representation.
The quadratic function \(g(x) = 2x^{2} + 5x - 3\) has its real zeros or x-intercepts where we set \(g(x) = 0\). These are the points that satisfy the equation \((2x - 1)(x + 3) = 0\) derived from our factoring.
- For \((2x - 1) = 0\), solving gives \(x = 0.5\).
- For \((x + 3) = 0\), solving gives \(x = -3\).
Thus, the real zeros are \(x = -3\) and \(x = 0.5\). These are also the x-values that you would see where the parabola crosses the x-axis on a graph. Real zeros provide insights into the nature of quadratic expressions and their graphical representation.
Other exercises in this chapter
Problem 23
In Exercises \(17-40,\) let \(f(x)=-x^{2}+x, g(x)=\frac{2}{x+1},\) and \(h(x)=-2 x+1 .\) Evaluate each of the following. $$(f-g)(2)$$
View solution Problem 23
Identify the underlying basic function, and use transformations of the basic function to sketch the graph of the given function. $$g(x)=-3(x+2)^{2}-4$$
View solution Problem 24
Solve the inequality by factoring. $$-3 x^{2} \leq-7 x-6$$
View solution Problem 24
Solve the rational equation. Check your solutions. $$\frac{1}{x^{2}}-\frac{7}{x}=18$$
View solution