Problem 50
Question
Solve each equation. $$ 9 x^{2}+7 x=2 $$
Step-by-Step Solution
Verified Answer
The solutions to the equation are \(x = \frac{2}{9}\) and \(x = -1\).
1Step 1: Move All Terms to One Side
First, we rewrite the equation by moving all terms to the left side to equal zero. The equation becomes: \[ 9x^2 + 7x - 2 = 0 \]
2Step 2: Identify Coefficients for the Quadratic Formula
The quadratic equation is in the form \(ax^2 + bx + c = 0\). Here, \(a = 9\), \(b = 7\), and \(c = -2\).
3Step 3: Apply the Quadratic Formula
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find \(x\). Substitute \(a = 9\), \(b = 7\), and \(c = -2\) into the formula.
4Step 4: Calculate the Discriminant
The discriminant \(\Delta\) is \(b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\): \[ \Delta = 7^2 - 4 \cdot 9 \cdot (-2) = 49 + 72 = 121 \]
5Step 5: Solve for x Using the Calculated Discriminant
Since the discriminant is 121, a perfect square, it indicates real and rational roots. Solve for \(x\) using the quadratic formula: \[ x = \frac{-7 \pm \sqrt{121}}{18} \] This simplifies to \[ x = \frac{-7 \pm 11}{18} \]
6Step 6: Find the Two Possible Solutions
Calculate the two possible solutions from step 5:- For \(x = \frac{-7 + 11}{18} = \frac{4}{18} = \frac{2}{9}\)- For \(x = \frac{-7 - 11}{18} = \frac{-18}{18} = -1\)
Key Concepts
Quadratic FormulaDiscriminantSolving EquationsReal and Rational Roots
Quadratic Formula
When faced with a quadratic equation like \(ax^2 + bx + c = 0\), the quadratic formula is your go-to tool for finding solutions. The formula is expressed as:
For example, in the equation \(9x^2 + 7x - 2 = 0\), we have \(a = 9\), \(b = 7\), and \(c = -2\). These values are inserted into the quadratic formula to calculate \(x\), providing us with the potential solutions.
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For example, in the equation \(9x^2 + 7x - 2 = 0\), we have \(a = 9\), \(b = 7\), and \(c = -2\). These values are inserted into the quadratic formula to calculate \(x\), providing us with the potential solutions.
Discriminant
The discriminant is a key component in the quadratic formula, found under the square root symbol as \(b^2 - 4ac\). It provides crucial information about the nature of the roots without needing to solve the entire equation.
Determining the discriminant can tell us:
Determining the discriminant can tell us:
- If \(\Delta > 0\), there are two distinct real roots.
- If \(\Delta = 0\), there is exactly one real root or a repeated root.
- If \(\Delta < 0\), the equation has no real roots; instead, it has two complex roots.
Solving Equations
Solving quadratic equations efficiently often involves transforming the equation into a form that can easily be handled by the quadratic formula. Begin by moving all terms to one side of the equation, aiming to achieve a structure such as \(ax^2 + bx + c = 0\).
Using the given quadratic formula, plug in the identified coefficients to resolve for \(x\). For example, from our equation, once we know \(a = 9\), \(b = 7\), and \(c = -2\), substituting these into the quadratic formula allows us to solve for \(x\) as follows:
Using the given quadratic formula, plug in the identified coefficients to resolve for \(x\). For example, from our equation, once we know \(a = 9\), \(b = 7\), and \(c = -2\), substituting these into the quadratic formula allows us to solve for \(x\) as follows:
- Calculate the discriminant \(\Delta = 49 + 72 = 121\).
- Evaluate \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Simplify to find the roots.
Real and Rational Roots
After solving the quadratic equation, you'll find the nature of its roots. Since we found that the discriminant \(\Delta = 121\) is a perfect square, the roots of this equation are real and rational.
Rational roots mean that both solutions can be expressed as fractions or whole numbers. In our given problem, the solutions turn out to be:
Rational roots mean that both solutions can be expressed as fractions or whole numbers. In our given problem, the solutions turn out to be:
- \(x = \frac{2}{9}\)
- \(x = -1\)
Other exercises in this chapter
Problem 50
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