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Chapter 11: Problem 65

Solve. Approximate the solutions to three decimal places. $$ z^{2}+0.84 z-0.4=0 $$

Short Answer

Expert verified
The solutions are approximately 0.339 and -1.179.

Step by step solution

01

Identify the quadratic equation format

The given equation is a quadratic equation in the form of \( az^{2} + bz + c = 0 \) where \( a = 1 \), \( b = 0.84 \), and \( c = -0.4 \).
02

Use the quadratic formula

To find the roots of the quadratic equation, use the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
03

Plug in the coefficients

Substitute \( a = 1 \), \( b = 0.84 \), and \( c = -0.4 \) into the formula: \( z = \frac{-0.84 \pm \sqrt{(0.84)^2 - 4(1)(-0.4)}}{2(1)} \).
04

Simplify inside the square root

Calculate the discriminant: \( (0.84)^2 - 4(1)(-0.4) = 0.7056 + 1.6 = 2.3056 \). Now, take the square root: \( \sqrt{2.3056} \approx 1.518 \).
05

Solve for the two possible values of z

Now solve for the two possible values using the values found: \( z_1 = \frac{-0.84 + 1.518}{2} \approx 0.339 \) and \( z_2 = \frac{-0.84 - 1.518}{2} \approx -1.179 \).
06

Approximate the solutions to three decimal places

The solutions to the quadratic equation, approximated to three decimal places, are \( z_1 \approx 0.339 \) and \( z_2 \approx -1.179 \).

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Key Concepts

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

Quadratic Formula
To solve quadratic equations of the form \( ax^2 + bx + c = 0 \), we use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
This formula provides solutions for any quadratic equation, as long as you know the coefficients \( a \), \( b \), and \( c \).

In our exercise:
  • \( a = 1 \)
  • \( b = 0.84 \)
  • \( c = -0.4 \)
Plugging these into the quadratic formula helps find the solutions for \( z \).
Discriminant
The discriminant in the quadratic formula is the part under the square root:

\( b^2 - 4ac \).

This value determines the nature of the roots:
  • If the discriminant is positive, there are two real and distinct solutions.
  • If it is zero, there is exactly one real solution.
  • If it is negative, the equation has two complex solutions.
In our example:
\[ (0.84)^2 - 4(1)(-0.4) = 0.7056 + 1.6 = 2.3056 \]
The discriminant is 2.3056, which is positive, meaning we have two real roots.
Approximating Solutions
After determining the discriminant and solving the quadratic equation, we need to approximate the solutions. This is done by calculating the square root and the exact decimal values up to the desired precision.

For our exercise:
The solution steps were:
\
  • \[ z_1 = \frac{-0.84 + \sqrt{2.3056}}{2} \approx \frac{-0.84 + 1.518}{2} \approx 0.339 \]
  • \[ z_2 = \frac{-0.84 - \sqrt{2.3056}}{2} \approx \frac{-0.84 - 1.518}{2} \approx -1.179 \]
These values are then rounded to three decimal places, providing the approximate roots for the quadratic equation: \( z_1 \approx 0.339 \) and \( z_2 \approx -1.179 \).

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Most popular questions from this chapter

To prepare for Section \(11.2,\) review evaluating expressions and simplifying radical expressions (Sections \(1.8,10.3\) and \(10.8)\) Simplify. \([10.3],[10.8]\) $$ \sqrt{-4} $$

For each of the following, write the equation of the parabola that has the shape of \(f(x)=2 x^{2}\) or \(g(x)=-2 x^{2}\) and has a maximum value or \(a\) minimum value at the specified point. Minimum: \((2,0)\)

Solve. $$ 7 x^{2}=21 $$

Review finding intercepts and completing the square (Sections \(3.3,6.2,6.3, \text { and } 11.1)\) Find the \(x\) -intercepts. $$ y=2 x^{2}-x-3 $$

Let \(f(t)=(t+6)^{2} .\) Find \(t\) such that \(f(t)=15\).
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