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Chapter 1: Problem 86

Find the real solutions, if any, of each equation. Use any method. $$ 2=y+6 y^{2} $$

Short Answer

Expert verified
The real solutions are \( y = \frac{1}{2} \) and \( y = -\frac{2}{3} \).

Step by step solution

01

Rewrite the equation in standard form

The given equation is \[ 2 = y + 6y^2 \]Rewrite it in the standard form of a quadratic equation by moving all terms to one side: \[ 6y^2 + y - 2 = 0 \]
02

Identify coefficients

Identify the coefficients from the standard form equation \[ 6y^2 + y - 2 = 0 \].Here, \(a = 6\), \(b = 1\), and \(c = -2\).
03

Use the quadratic formula

The quadratic formula is \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].Substitute \(a = 6\), \(b = 1\), and \(c = -2\) into the quadratic formula.
04

Calculate the discriminant

The discriminant is given by \( b^2 - 4ac \).Substitute the values to find the discriminant: \[ 1^2 - 4(6)(-2) = 1 + 48 = 49 \]
05

Solve for y

Since the discriminant is 49, which is positive, there are two real solutions:\[ y = \frac{-1 \pm \sqrt{49}}{12} = \frac{-1 \pm 7}{12} \]This gives us: \[ y_1 = \frac{6}{12} = \frac{1}{2} \]\[ y_2 = \frac{-8}{12} = -\frac{2}{3} \]

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Key Concepts

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

Real Solutions
In mathematics, the term 'real solutions' refers to solutions of an equation that are real numbers鈥攏ot imaginary or complex numbers. When solving quadratic equations (equations of the form \(ax^2 + bx + c = 0\)), we are typically interested in identifying whether the solutions are real or not. This can be determined by analyzing the discriminant, the part under the square root in the quadratic formula. If the discriminant is positive, the quadratic equation has two distinct real solutions. If it is zero, there is exactly one real solution. However, if the discriminant is negative, the quadratic equation has no real solutions but two complex solutions.
Quadratic Formula
The quadratic formula is a powerful tool used for solving quadratic equations. It can be used when factoring is difficult or not possible. The formula is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
To use the quadratic formula, follow these steps:
  • First, rewrite the quadratic equation in its standard form where \a, b,\ and \c\ are the coefficients of the terms \(ax^2 + bx + c = 0\).
  • Identify \a, b,\ and \c\ in the equation.
  • Substitute these values into the quadratic formula.
  • Calculate the discriminant (the expression under the square root).
  • Finally, solve for \y\ by simplifying the expression.
This method guarantees finding all real (and complex, if applicable) solutions to any quadratic equation.
Discriminant
The discriminant is a specific part of the quadratic formula under the square root symbol. It is given by: \[ b^2 - 4ac \].
The value of the discriminant reveals important information about the nature of the roots of a quadratic equation. Depending on the value of the discriminant:
  • If \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
  • If \( b^2 - 4ac = 0 \), the equation has exactly one real solution (a repeated or double root).
  • If \( b^2 - 4ac < 0 \), the equation has two complex solutions (no real solutions).
Understanding the discriminant helps us to predict the type of solutions we can expect without actually solving the quadratic equation.

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