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Chapter 4: Problem 220

Use the quadratic formula to find the solutions to \(x^{2}+x-1=\) \(0,\) and use that information to factor \(x^{2}+x-1 .\)

Short Answer

Expert verified
\(\frac{-1 + \sqrt{5}}{2}\), \(\frac{-1 - \sqrt{5}}{2}\), and factored as \((x - \frac{-1 + \sqrt{5}}{2})(x - \frac{-1 - \sqrt{5}}{2})\).

Step by step solution

01

Identify coefficients

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

Write the quadratic formula

The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
03

Substitute coefficients into the formula

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

Simplify inside the square root

Simplify the expression inside the square root: \(x = \frac{-1 \pm \sqrt{1 + 4}}{2}\) \(x = \frac{-1 \pm \sqrt{5}}{2}\).
05

Solve for the values of x

Solve for \(x\) using the quadratic formula: \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\).
06

Write the factors

The solutions \(x_1\) and \(x_2\) give the factors of the quadratic expression: \((x - x_1)(x - x_2) = 0\). Thus, the factors are: \(\left(x - \frac{-1 + \sqrt{5}}{2}\right)\left(x - \frac{-1 - \sqrt{5}}{2}\right)\).

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Key Concepts

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

quadratic formula
Solving quadratic equations involves finding the values of x that satisfy the equation. The quadratic formula is a crucial tool for this purpose.
The formula can solve any quadratic equation of the form: \( ax^2 + bx + c = 0 \).
Here is the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
To use it effectively:
  • Identify the coefficients a, b, and c in your equation.
  • Substitute these values into the quadratic formula.
  • Simplify the expression under the square root, known as the discriminant \((b^2 - 4ac)\).
  • Solve for x by simplifying and splitting into two solutions using the \( \pm \).
Using our example \( x^2 + x - 1 = 0 \), determining the coefficients gives: \(a = 1\), \(b = 1\), and \(c = -1\).
Substitute these values to get \( x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-1)}}{2(1)} \), which simplifies to \( x = \frac{-1 \pm \sqrt{5}}{2} \).
This results in two solutions: \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \).
factoring quadratics
Factoring quadratic equations is another technique to solve them.
A quadratic can often be expressed as a product of two binomials: \((x - x_1)(x - x_2) = 0\).
To factorize using the roots found from the quadratic formula:
  • First find the roots of the equation using the quadratic formula.
  • Use these roots to express the quadratic equation in its factored form.
From our example \( x^2 + x - 1 = 0 \), we found the roots to be \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \).
Therefore, the factored form is: \( \left(x - \frac{-1 + \sqrt{5}}{2}\right) \left(x - \frac{-1 - \sqrt{5}}{2}\right) \).
This expresses the quadratic as a product of linear factors, using its roots.
mathematical problem solving
Effective problem-solving requires a systematic approach.
For quadratic equations, this often involves:
  • Comprehending the problem and identifying the type of equation at hand.
  • Choosing the right methods, like the quadratic formula or factoring, based on the problem's structure.
  • Applying the method step-by-step without skipping steps.
  • Double-checking each calculation to avoid errors.
Let's break down our problem-solving approach for the given equation:
Start by recognizing it's a quadratic equation. Decide that the quadratic formula suits this case.
Identify the coefficients (i.e., \(a = 1\), \(b = 1\), \(c = -1\)) and substitute them into the formula.
Simplify step-by-step and solve for roots (\( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \)).
Finally, reformat the roots into a factored quadratic for the solution.
algebra
Algebra involves working with symbols and following specific operations to solve equations.
It's fundamental for understanding higher-level math and various fields like science and engineering.
When dealing with quadratic equations, algebraic manipulation is key.
Let's apply algebra to our example:
We start with the equation \( x^2 + x - 1 = 0 \).
Recognize that it鈥檚 in the standard form \( ax^2 + bx + c = 0 \).
To solve it, we:
  • Identify a, b, and c.
  • Use algebraic principles to manipulate and substitute into the quadratic formula.
  • Perform operations to simplify the roots.
  • Express the solution in terms of linear factors using the roots found.
By following algebraic rules, we derive the solution and fully understand the problem context.
Master algebra to excel in quadratic equations and beyond.

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Most popular questions from this chapter

In Problem 217 you may have simply guessed at values of \(c\) and \(d\), or you may have solved a system of equations in the two unknowns \(c\) and \(d\). Given constants \(a, b, r_{1},\) and \(r_{2}\) (with \(\left.r_{1} \neq r_{2}\right)\), write down a system of equations we can solve for \(c\) and \(d\) to write $$ \frac{a x+b}{\left(x-r_{1}\right)\left(x-r_{2}\right)}=\frac{c}{x-r_{1}}+\frac{d}{x-r_{2}} $$

If we have five identical pennies, five identical nickels, five identical dimes, and five identical quarters, give the picture enumerator for the combinations of coins we can form and convert it to a generating function for the number of ways to make \(k\) cents with the coins we have. Do the same thing assuming we have an unlimited supply of pennies, nickels, dimes, and quarters. (w)

Use generating functions to explain why the number of partitions of an integer in which each part is used an even number of times equals the generating function for the number of partitions of an integer in which each part is even. (h)

In the polynomial \(\left(a_{0}+a_{1} x+a_{2} x^{2}\right)\left(b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}\right)\), what is the coefficient of \(x^{2} ?\) What is the coefficient of \(x^{4} ?\)

Suppose we start (at the end of month 0 ) with 10 pairs of baby rabbits, and that after baby rabbits mature for one month they begin to reproduce, with each pair producing two new pairs at the end of each month afterwards. Suppose further that over the time we observe the rabbits, none die. Let \(a_{n}\) be the number of rabbits we have at the end of month \(n\). Show that \(a_{n}=a_{n-1}+2 a_{n-2}\). This is an example of a second order linear recurrence with constant coefficients. Using a method similar to that of Problem 211 , show that $$ \sum_{i=0}^{\infty} a_{i} x^{i}=\frac{10}{1-x-2 x^{2}} $$ This gives us the generating function for the sequence \(a_{i}\) giving the population in month \(i\); shortly we shall see a method for converting this to a solution to the recurrence.
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