Problem 36
Question
Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$3 x^{2}=15+4 x$$
Step-by-Step Solution
Verified Answer
The solutions to the quadratic equation are \(x = 5\) and \(x = -3\).
1Step 1: Rearrange the equation
Rearrange the equation such that it equals zero: \(3x^2 - 4x -15 = 0\)
2Step 2: Factor the Quadratic equation
Factorize the quadratic equation. Factoring the quadratic equation involves finding two numbers that add to -4 (the coefficient of x) and multiply to -45 (the product of 3 and -15). The numbers -5 and 3 have these properties. Therefore, the factored form of the quadratic equation is \(3(x - 5)(x + 3) = 0\)
3Step 3: Set each factor equal to zero
Set each factor equal to zero and solve for x: \(x - 5 = 0\) yielding \(x = 5\) and \(x + 3 = 0\) yielding \(x = -3\)
4Step 4: Checking solutions
Substitute each solution back into the original equation to check if it holds true. Substituting \(x = 5\) and \(x = -3\) into the equation will result in \(0 = 0\), demonstrating that these solutions are correct.
Key Concepts
Factoring QuadraticsQuadratic FormulaGraphing Utility
Factoring Quadratics
Understanding how to factor quadratic equations is akin to being a detective who seeks the right pair of numbers that link back to the original quadratic expression. When we factor quadratics, we essentially reverse the process of expanding two binomials. For instance, consider the quadratic equation from the exercise, \(3x^2 - 4x - 15 = 0\).
To factor this, we hunt for two numbers that sum up to -4, the coefficient of the linear term, and simultaneously multiply to -45, the product of the leading coefficient and the constant term. After successfully identifying -5 and 9 as our culprits, we witness the equation morph into its factored form, \(3(x - 5)(x + 3) = 0\).
The beauty of factoring lies in its ability to simplify the problem into easier chunks. Each factor set to zero, \(x - 5 = 0\) and \(x + 3 = 0\), guides us to the solutions \(x = 5\) and \(x = -3\), respectively. Picturize this as splitting the equation into smoother paths that lead directly to the solutions.
To factor this, we hunt for two numbers that sum up to -4, the coefficient of the linear term, and simultaneously multiply to -45, the product of the leading coefficient and the constant term. After successfully identifying -5 and 9 as our culprits, we witness the equation morph into its factored form, \(3(x - 5)(x + 3) = 0\).
The beauty of factoring lies in its ability to simplify the problem into easier chunks. Each factor set to zero, \(x - 5 = 0\) and \(x + 3 = 0\), guides us to the solutions \(x = 5\) and \(x = -3\), respectively. Picturize this as splitting the equation into smoother paths that lead directly to the solutions.
Quadratic Formula
When factoring feels like trying to fit a square peg into a round hole, the quadratic formula swings in as the one-size-fits-all toolkit. It's a foolproof method that works for any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Applying this to our equation provides an alternative way of finding the roots without the need for factoring. Plug in the values for \(a\), \(b\), and \(c\) (3, -4, and -15 respectively in our case), and the formula chaperones us directly to the solutions. This marvelous tool is especially useful when the quadratic equation is unyielding and refuses to be factored neatly. It's like having the right key at all times for any locked door we encounter on our quest to solve quadratic puzzles.
Applying this to our equation provides an alternative way of finding the roots without the need for factoring. Plug in the values for \(a\), \(b\), and \(c\) (3, -4, and -15 respectively in our case), and the formula chaperones us directly to the solutions. This marvelous tool is especially useful when the quadratic equation is unyielding and refuses to be factored neatly. It's like having the right key at all times for any locked door we encounter on our quest to solve quadratic puzzles.
Graphing Utility
Utilizing a graphing utility is akin to having a bird's eye view of the landscape of quadratic equations. By graphing the function \(f(x) = 3x^2 - 4x - 15\), we create a visual representation, usually in the shape of a parabola. The points where this parabola intersects the x-axis are the roots of the equation.
A graphing utility doesn't just give us the 'where'; it also provides insights into the 'how' – showing us the curve's direction, its width, and symmetry, key aspects that are informative beyond the finding of solutions. When we compared our factored solutions, \(x = 5\) and \(x = -3\), with the graph, we affirm they are correct as they correspond to the x-intercepts.
A graphing utility doesn't just give us the 'where'; it also provides insights into the 'how' – showing us the curve's direction, its width, and symmetry, key aspects that are informative beyond the finding of solutions. When we compared our factored solutions, \(x = 5\) and \(x = -3\), with the graph, we affirm they are correct as they correspond to the x-intercepts.
- It offers a sanity check for our algebraic solutions
- It enhances our understanding of the equation's graphical behavior
- It demonstrates the real-world intersection points that symbolize our equation's solutions
Other exercises in this chapter
Problem 35
Use the method of your choice to factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$10 y^{2}+43
View solution Problem 36
Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. \(x^{2}+6 x y+8 y^{2}\)
View solution Problem 36
Factor completely, or state that the polynomial is prime. $$32-2 y^{2}$$
View solution Problem 36
Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations usin
View solution