Problem 116
Question
In Exercises \(115-122,\) find all values of \(x\) satisfying the given conditions. $$ y=5 x^{2}+3 x \text { and } y=2 $$
Step-by-Step Solution
Verified Answer
The possible values for \(x\) that satisfy the equation are \(x = 0.4\) and \(x = -1\)
1Step 1: Substitute Given 'y' Value into the Equation
Replace \(y\) in the equation \(y = 5x^2 + 3x\) with the given value of \(2\). This gives: \(2 = 5x^2 + 3x\).
2Step 2: Re-arrange the Equation and Set it Equal to Zero
In order to apply the quadratic formula, our equation should be in the form \(ax^2 + bx + c = 0\). By subtracting \(2\) from both sides of the equation, we get \(5x^2 + 3x - 2 = 0\).
3Step 3: Apply the Quadratic Formula to Solve for 'x'
Write down the quadratic formula, \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\), and then substitute \(a = 5\), \(b = 3\), and \(c = -2\) into the formula. The solution is \(x = [-3 ± sqrt((3)^2 - 4*5*(-2))] / (2*5)\). Simplify the equation to get \(x = [-3 ± sqrt(9+40)] / 10 = [-3 ± sqrt(49)] / 10\). Therefore, \(x = [-3 ± 7] / 10 \). This gives us two solutions: \(x = 4/10 = 0.4\) and \(x = -10/10 = -1\).
Key Concepts
Quadratic FormulaZero Product PropertySubstitute Values in Equations
Quadratic Formula
The quadratic formula is a powerful tool that allows us to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). To understand why this formula works, think of a quadratic equation as a curve on a graph representing all possible solutions. The solutions are where this curve crosses the x-axis, these points are also known as the roots of the equation.
When we apply the quadratic formula, \( x = [-b \pm \sqrt{b^2 - 4ac}] / (2a) \), we're essentially calculating these intersection points. The symbols \( \pm \) indicate that we will get two solutions, which correspond to the two places where our curve crosses the x-axis. The discriminant, \( b^2-4ac \), determines whether the roots are real numbers (if its value is positive), a single real number (if zero), or complex numbers (if negative).
In the given exercise, using the quadratic formula led to finding the two values of \( x \) that satisfy the equation \( y = 5x^2 + 3x \) when \( y = 2 \). The formula offered a straightforward path to calculating these roots without having to rely on graphing or other methods of solving quadratics, demonstrating its utility and significance in algebra.
When we apply the quadratic formula, \( x = [-b \pm \sqrt{b^2 - 4ac}] / (2a) \), we're essentially calculating these intersection points. The symbols \( \pm \) indicate that we will get two solutions, which correspond to the two places where our curve crosses the x-axis. The discriminant, \( b^2-4ac \), determines whether the roots are real numbers (if its value is positive), a single real number (if zero), or complex numbers (if negative).
In the given exercise, using the quadratic formula led to finding the two values of \( x \) that satisfy the equation \( y = 5x^2 + 3x \) when \( y = 2 \). The formula offered a straightforward path to calculating these roots without having to rely on graphing or other methods of solving quadratics, demonstrating its utility and significance in algebra.
Zero Product Property
The zero product property states that if the product of two terms is zero, then at least one of the terms must be zero. This property is particularly useful when solving quadratic equations because after factoring, we can set each factor equal to zero to find the roots of the equation.
Consider a factored quadratic equation \( (x - p)(x - q) = 0 \). According to the zero product property, either \( x - p = 0 \) or \( x - q = 0 \), leading to the solutions \( x = p \) or \( x = q \). While this approach wasn't directly applied in our original exercise, as the quadratic formula was used instead, understanding the zero product property is foundational to other methods of solving quadratics, such as factoring. This property would have come into play if the quadratic \( 5x^2 + 3x - 2 = 0 \) had been easy to factor into a form such as \( (x - a)(x - b) = 0 \) where \( a \) and \( b \) are the roots of the equation.
Consider a factored quadratic equation \( (x - p)(x - q) = 0 \). According to the zero product property, either \( x - p = 0 \) or \( x - q = 0 \), leading to the solutions \( x = p \) or \( x = q \). While this approach wasn't directly applied in our original exercise, as the quadratic formula was used instead, understanding the zero product property is foundational to other methods of solving quadratics, such as factoring. This property would have come into play if the quadratic \( 5x^2 + 3x - 2 = 0 \) had been easy to factor into a form such as \( (x - a)(x - b) = 0 \) where \( a \) and \( b \) are the roots of the equation.
Substitute Values in Equations
When solving equations, substituting specific values into equations is a fundamental skill. It allows us to see how altering one variable affects another and moves us closer to finding a solution.
In the context of our exercise, we started with two equivalent expressions for \( y \): the quadratic equation \( y = 5x^2 + 3x \) and the constant \( y = 2 \). By substituting \( 2 \) for \( y \) in the quadratic equation, we effectively transformed the problem into one where we could apply the quadratic formula.
In the context of our exercise, we started with two equivalent expressions for \( y \): the quadratic equation \( y = 5x^2 + 3x \) and the constant \( y = 2 \). By substituting \( 2 \) for \( y \) in the quadratic equation, we effectively transformed the problem into one where we could apply the quadratic formula.
Why is substitution important?
- Provides a direct method for solving equations.
- Helps in simplifying and re-structuring equations into solvable forms.
- It is a strategy that can be applied across various types of equations, not just quadratics.
Other exercises in this chapter
Problem 115
In Exercises \(115-122,\) find all values of \(x\) satisfying the given conditions. $$ y=2 x^{2}-3 x \text { and } y=2 $$
View solution Problem 115
Without actually solving the equation, give a general description of how to solve \(x^{3}-5 x^{2}-x+5=0\)
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In solving \(\sqrt{3 x+4}-\sqrt{2 x+4}=2,\) why is it a good idea to isolate a radical term? What if we don't do this and simply square each side? Describe what
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Use your graphing utility to enter each side of the equation separately under \(y_{1}\) and \(y_{2} .\) Then use the utility's TABLE or GRAPH feature to solve t
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