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Chapter 11: Problem 12

Solve the equation by using the quadratic formula where appropriate. $$y^{2}+2 y-5=0$$

Short Answer

Expert verified
The solutions are \[ x = -1 + \sqrt{6} \] and \[ x = -1 - \sqrt{6} \].

Step by step solution

01

- Identify coefficients

Recognize the standard form of a quadratic equation: \[ ax^2 + bx + c = 0 \]In this equation, the coefficients are: \[ a = 1, b = 2, c = -5 \]
02

- Recall the quadratic formula

The quadratic formula for solving \[ ax^2 + bx + c = 0 \] is: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
03

- Substitute the coefficients into the formula

Substitute \[ a = 1, b = 2, c = -5 \] into the quadratic formula: \[ x = \frac{{-2 \pm \sqrt{{2^2 - 4(1)(-5)}}}}{2(1)} \]
04

- Simplify inside the square root

Calculate the value inside the square root: \[ x = \frac{{-2 \pm \sqrt{{4 + 20}}}}{2} \] \[ x = \frac{{-2 \pm \sqrt{{24}}}}{2} \]
05

- Simplify the square root and the fraction

Simplify the square root of 24 and the fraction: \[ \sqrt{24} = 2 \sqrt{6} \] Substitute back: \[ x = \frac{{-2 \pm 2\sqrt{6}}}{2} \]Simplify the fraction: \[ x = -1 \pm \sqrt{6} \]

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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 to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). This formula is derived from the process of completing the square and is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. The term \(b^2 - 4ac\) inside the square root is called the discriminant, and it determines the nature of the roots:
  • If \(b^2 - 4ac > 0\), there are two distinct real roots.
  • If \(b^2 - 4ac = 0\), there is one real root (also called a repeated or double root).
  • If \(b^2 - 4ac < 0\), there are no real roots; instead, the roots are complex.
Applying the quadratic formula involves substituting the coefficients into the formula and simplifying step-by-step.
coefficients in quadratic equations
Understanding the role of coefficients in quadratic equations is crucial. In the standard form \(ax^2 + bx + c = 0\), the coefficients are:
- \(a\) is the coefficient of \(x^2\).
- \(b\) is the coefficient of \(x\).
- \(c\) is the constant term.In our example, \(y^2 + 2y - 5 = 0\), the coefficients are identified as:
  • \(a = 1\), which is the coefficient of \(y^2\).
  • \(b = 2\), which is the coefficient of \(y\).
  • \(c = -5\), which is the constant term.
These values are essential for plugging into the quadratic formula and solving the equation.
simplifying square roots
Simplifying square roots is an important step when using the quadratic formula. In the equation provided (\( y^2 + 2y - 5 = 0 \)), after substitution and simplification, we reached:
\( x = \frac{-2 \pm \sqrt{24}}{2} \)
To simplify \(\sqrt{24}\):
  • First, recognize that 24 can be factored into 4 and 6: \(\sqrt{24} = \sqrt{4 \cdot 6}\)
  • Next, split the square root: \(\sqrt{4} \cdot \sqrt{6} = 2\sqrt{6}\)
Substitute back into the quadratic formula:
\( x = \frac{-2 \pm 2\sqrt{6}}{2} \)
Finally, simplify the fraction by dividing both terms in the numerator by the denominator:
\( x = -1 \pm \sqrt{6} \). By breaking down the process, it becomes much clearer on how to simplify square roots and solve quadratic equations.

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Most popular questions from this chapter

$$\text { In Exercises } 97-100, \text { solve the system of equations algebraically.}$$ $$\left\\{\begin{array}{l} 2 x+y=6 \\ 3 x-2 y=23 \end{array}\right.$$

$$\text { In Exercises } 97-100, \text { solve the system of equations algebraically.}$$ $$\left\\{\begin{array}{l} 6 x-4 y=9 \\ 3 x-2 y=2 \end{array}\right.$$

In Exercises \(65-74\), solve the given equation. For quadratic equations, choose either the factoring method or the square root method, whichever you think is the easier to use. $$\frac{1}{x}+\frac{6}{x+2}=2$$

Sketch the graph of the given equation. Find the intercepts; approximate to the nearest tenth where necessary. $$y=(x-1)^{2}$$

In Exercises \(85-94\), solve the equations using the square root method. Round off your answers to the nearest hundredth. $$t^{2}=21$$
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