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Chapter 10: Problem 100

For Exercises 94 to \(101,\) solve. $$(2 z-3)(z+5)=(z+1)(z+3)$$

Short Answer

Expert verified
The solutions for the given quadratic equation are \(z = -6\) and \(z = 3\)

Step by step solution

01

Expand both sides of the equation

Expand the brackets on both sides to get: \(2z^2 + 7z - 15 = z^2 + 4z + 3\)
02

Simplify to get standard quadratic equation form

To obtain the standard form, subtract \(z^2 + 4z + 3\) from each side: \(2z^2 - z^2 + 7z - 4z - 15 - 3 = 0\), simplifying we get \(z^2 + 3z - 18 = 0\)
03

Factorize the quadratic equation

Factorize the quadratic equation to find the roots (i.e., when the equation equals zero). The factorized form is \(z+6)(z-3)=0\)
04

Solve for z

By setting each factor equal to zero and solving for z, we obtain \(z = -6, 3\)

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Key Concepts

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

Factoring Quadratics
Factoring quadratics is a key process in algebra. It involves rewriting a quadratic equation in the form of
  • The goal is to express the equation as a product of two binomials.
  • This is especially useful because it allows us to find the roots of the equation more easily.
A quadratic equation in standard form looks like:
\( ax^2 + bx + c = 0 \). The process of factoring seeks two numbers that multiply to \( c \) (the constant term) and add up to \( b \) (the coefficient of the linear term).
For instance, considering \( z^2 + 3z - 18 = 0 \), we look for two numbers that multiply to \(-18\) (product term) and add to \(3\) (sum term).
These numbers are +6 and -3, leading to the factors \( (z + 6)(z - 3) = 0 \).
Once factored, solving the equation becomes straightforward; set each factor to zero and solve for the variable.
Polynomial Expansion
Polynomial expansion is the process of removing parentheses by using the distributive property to multiply out expressions.
  • This is the opposite of factoring.
  • It often involves using the distributive property multiple times or applying identities like \((a + b)^2 = a^2 + 2ab + b^2\).
In our exercise, both sides of the equation were expanded from their original forms:
The left side, \((2z-3)(z+5)\), expands to \(2z^2 + 10z - 3z - 15\), which simplifies to \(2z^2 + 7z - 15\).
The right side, \((z+1)(z+3)\), expands to \(z^2 + 3z + z + 3\), simplifying to \(z^2 + 4z + 3\).
This process is crucial because it transforms a balanced equation into a polynomial form, allowing us to identify terms that enable further simplification.
Roots of Equations
Finding roots of equations is one of the primary objectives when dealing with polynomials.
  • The roots are the values of the variable that make the equation true.
  • In terms of a graph, these are the points where the curve intersects the x-axis.
After factoring the quadratic equation \( z^2 + 3z - 18 = 0 \) into \( (z + 6)(z - 3) = 0 \), the roots become apparent.
Setting each factor equal to zero gives us the possible solutions for \( z \):
1. From \( z + 6 = 0 \), we find \( z = -6 \).
2. From \( z - 3 = 0 \), we find \( z = 3 \).
Thus, the roots of the equation are \( z = -6 \) and \( z = 3 \), showing the values where the original quadratic equals zero, indicating points of intersection on a graph with the x-axis.

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Most popular questions from this chapter

The area of a rectangle is \(\left(3 x^{2}+x-2\right) \mathrm{ft}^{2} .\) Find the dimensions of the rectangle in terms of the variable \(x .\) Given that \(x>0,\) specify the dimension that is the length and the dimension that is the width. Can \(x\) be negative? Can \(x=0 ?\) Explain your answers. $$A=3 x^{2}+x-2$$

Factor by grouping. $$4 y^{2}-11 y z+6 z^{2}$$

Find all integers \(k\) such that the trinomial can be factored over the integers. $$2 x^{2}+k x-5$$

Factor by grouping. $$18 y^{2}-39 y+20$$

For Exercises 144 to \(149,\) determine the positive integer values of \(k\) for which the polynomial is factorable over the integers. $$c^{2}-7 c+k$$
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