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Chapter 11: Problem 14

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 2 x^{2}+3 x-1=0 $$

Short Answer

Expert verified
The solutions are \(x = \frac{-3 + \sqrt{17}}{4}\) and \(x = \frac{-3 - \sqrt{17}}{4}\).

Step by step solution

01

- Identify the coefficients

For the quadratic equation in the form of \(ax^2 + bx + c = 0\), identify the coefficients \(a\), \(b\), and \(c\). In this case, \(a = 2\), \(b = 3\), and \(c = -1\).
02

- Write the quadratic formula

Recall the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Substitute the identified coefficients \(a\), \(b\), and \(c\) into the quadratic formula.
03

- Calculate the discriminant

Compute the discriminant using the formula \(b^2 - 4ac\). For our coefficients, \(3^2 - 4(2)(-1) = 9 + 8 = 17\).
04

- Substitute into the quadratic formula

Substitute the coefficients and the discriminant value back into the quadratic formula: \[x = \frac{-3 \pm \sqrt{17}}{4}\].
05

- Compute the solutions

Use the quadratic formula to find the two solutions: \[x = \frac{-3 + \sqrt{17}}{4}\] and \[x = \frac{-3 - \sqrt{17}}{4}\]. These are the solutions to the quadratic equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving quadratic equations
Quadratic equations are polynomials of the form \[ax^2 + bx + c = 0\], where \(a\), \(b\), and \(c\) are constants. These equations can have one, two, or no real solutions.
The quadratic formula is a powerful tool to find these solutions. It is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
This formula might look complicated at first, but breaking it down step-by-step makes it easier.

  • Identify the coefficients \(a\), \(b\), and \(c\).

  • Compute the discriminant \(b^2 - 4ac\).

  • Substitute the values into the formula.

  • Simplify to get the solutions.


In the given equation, \[2x^2 + 3x - 1 = 0\], we identified \(a = 2\), \(b = 3\), and \(c = -1\).
Next, we calculated the discriminant, substituted everything into the formula, and found the solutions.
discriminant
The discriminant is a key part of the quadratic formula. It helps in determining the nature and number of the solutions.
The discriminant \(D\) is represented as \(D = b^2 - 4ac\).

  • If \(D > 0\), there are two distinct real solutions.

  • If \(D = 0\), there is exactly one real solution.

  • If \(D < 0\), there are no real solutions, only complex ones.


In the given example, we calculated the discriminant as follows: \[3^2 - 4(2)(-1) = 9 + 8 = 17\].
Since \(17 > 0\), we know there are two distinct real solutions for our equation. This understanding makes it easier to predict the number of solutions before solving the equation fully.
coefficients
Coefficients in a quadratic equation are the numbers in front of the variables \(x^2\), \(x\), and the constant term.
They are crucial for solving the equation using the quadratic formula.
  • In the standard form \(ax^2 + bx + c = 0\), \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term.

  • These coefficients help in setting up the quadratic formula.

  • We use them to calculate the discriminant.

  • Finally, they are substituted back into the quadratic formula to find the solutions.


In our example, we have \(a = 2\), \(b = 3\), and \(c = -1\).
Identifying these correctly allowed us to proceed with solving the equation accurately using the quadratic formula.

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Most popular questions from this chapter

The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(2 x^{4}+x^{2}-3=0\)

Solve using the square root property. Simplify all radicals. $$ 2 x^{2}=9 $$

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (2 x-1)(8 x-4)=-1 $$

Solve each equation by completing the square. Give (a) exact solutions and (b) solutions rounded to the nearest thousandth. \(3 r^{2}-2=6 r+3\)

Solve using the square root property. Simplify all radicals. $$ \left(x-\frac{1}{5}\right)^{2}=\frac{16}{25} $$
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