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Magazine Chapter 6: Problem 120
Solve each equation. $$5 x^{2}=6-13 x$$
Short Answer
Expert verified
The solutions are \(x = 0.4\) and \(x = -3\).
Step by step solution
01
Write the equation in standard form
First, we need to write the given equation in the standard form of a quadratic equation, which is \(ax^2 + bx + c = 0\). To do this, move all the terms to one side of the equation. Start with the given equation: \[5x^2 = 6 - 13x\]Subtract 6 and add 13x from both sides to obtain:\[5x^2 + 13x - 6 = 0\]
02
Identify the coefficients
Now that the equation is in standard form, identify the coefficients \(a\), \(b\), and \(c\). For the equation \(5x^2 + 13x - 6 = 0\), the coefficients are:- \(a = 5\)- \(b = 13\)- \(c = -6\)
03
Apply the quadratic formula
To solve for \(x\), use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute the values of \(a\), \(b\), and \(c\) into the formula:\[x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 5 \cdot (-6)}}{2 \cdot 5}\]
04
Calculate the discriminant
Calculate the value inside the square root, known as the discriminant:\[b^2 - 4ac = 13^2 - 4 \cdot 5 \cdot (-6)\]\[= 169 + 120 = 289\]
05
Solve for x using the quadratic formula
Now that we have the discriminant, substitute it back into the quadratic formula:\[x = \frac{-13 \pm \sqrt{289}}{10}\]Since \(\sqrt{289} = 17\), substitute back to find:\[x = \frac{-13 \pm 17}{10}\]
06
Calculate the two possible solutions for x
Finally, solve for \(x\) by calculating the two solutions:1. \(x = \frac{-13 + 17}{10} = \frac{4}{10} = 0.4\)2. \(x = \frac{-13 - 17}{10} = \frac{-30}{10} = -3\)Therefore, the solutions are \(x = 0.4\) and \(x = -3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations. It is used when factoring is complex or impossible. The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where:
- \(a\), \(b\), and \(c\) are the coefficients from the standard form of the quadratic equation \(ax^2 + bx + c = 0\).
- \(\pm\) indicates that there are typically two solutions.
Discriminant
The discriminant is the part of the quadratic formula under the square root, \(b^2 - 4ac\). It plays an essential role in understanding the nature of the roots of a quadratic equation.Here's why the discriminant matters:
- If the discriminant (\(b^2 - 4ac\)) is positive, the quadratic equation has two distinct real solutions. Think of these as two separate points where the parabola crosses the x-axis.
- If it is zero, there is exactly one real solution. This means the parabola just touches the x-axis at that point, called a repeated or double root.
- If the discriminant is negative, there are no real solutions; instead, the solutions are complex numbers. In this case, the parabola does not cross the x-axis at all.
Standard Form of Quadratic Equation
To solve a quadratic equation using methods like the quadratic formula, it's crucial to first convert it to the standard form, which is:\[ ax^2 + bx + c = 0 \]In this form:
- \(a\) is the coefficient of \(x^2\).
- \(b\) is the coefficient of \(x\).
- \(c\) is the constant term.
