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Chapter 1: Problem 53

Find the real solutions of each equation. $$ 2(s+1)^{2}-5(s+1)=3 $$

Short Answer

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

Step by step solution

01

Let substitution

Introduce a substitution to simplify the equation. Let \[ u = s + 1 \]. This transforms the equation into:\[ 2u^2 - 5u = 3 \]
02

Rewrite the equation

Rearrange the equation to set it to zero:\[ 2u^2 - 5u - 3 = 0 \]
03

Solve the quadratic equation

Use the quadratic formula \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 2 \), \( b = -5 \), and \( c = -3 \). Plug these values in:\[ u = \frac{5 \pm \sqrt{25 + 24}}{4} \]\[ u = \frac{5 \pm \sqrt{49}}{4} \]\[ u = \frac{5 \pm 7}{4} \]So, we get two solutions for \( u \):\[ u = 3 \] and \[ u = -\frac{1}{2} \]
04

Substitute back to find s

Recall \[ u = s + 1 \]. Substitute back the values of \( u \):For \( u = 3 \):\[ 3 = s + 1 \]\[ s = 2 \]For \( u = -\frac{1}{2} \):\[ -\frac{1}{2} = s + 1 \]\[ s = -\frac{3}{2} \]

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Key Concepts

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

Substitution Method
The substitution method is a helpful technique to simplify complex equations. In this exercise, we start by substituting a part of the given equation with a new variable. This reduces the complexity and makes solving easier.
For example, we have the equation:
\( 2(s+1)^2-5(s+1)=3\)
We introduce a substitution to simplify it. Let .
With this substitution, the original equation becomes:
\(2u^2 - 5u = 3\)
This new equation is simpler to work with compared to the original one. After transforming the equation, we can solve for u first and then revert to the original variable.
The substitution method not only simplifies the calculations but also clarifies the structure of the original equation.
Quadratic Formula
Quadratic equations come in the standard form: \(ax^2 + bx + c = 0\).
To solve these equations, we use the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
In our transformed equation:
\(2u^2 - 5u - 3 = 0\),
here,聽 \(a = 2\), \(b = -5\), and \(c = -3\).
Applying these values into the quadratic formula, we get:
\(u = \frac{5 \pm \sqrt{25 + 24}}{4}\)
\(u = \frac{5 \pm \sqrt{49}}{4}\)
\(u = \frac{5 \pm 7}{4}\)
This results in two solutions:
\(u = 3\) and聽 \(u = -\frac{1}{2}\).
The quadratic formula allows us to find the roots of the quadratic equation accurately and efficiently.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations to isolate variables and solve them. This includes steps like addition, subtraction, multiplication, and division.
In this exercise, after using substitution, we have:
\(2u^2 - 5u - 3 = 0\).
We manipulated the terms to get the standard form of a quadratic equation.
Next, after solving for using the quadratic formula, we obtained and .
Finally, we reversed the substitution to find the original variable . For u = 3:
\(3 = s + 1\)
\(s = 2\),
and for u = -\frac{1}{2}:
\(-\frac{1}{2} = s + 1\)
\(s = -\frac{3}{2}\).
Algebraic manipulation is essential for moving terms around, simplifying expressions, and ultimately solving equations.

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Most popular questions from this chapter

Find the real solutions, if any, of each equation. Use any method. $$ x^{2}-6=0 $$

Solve: \(\sqrt{3 x+5}-\sqrt{x-2}=\sqrt{x+3}\)

Pythagorean Theorem How many right triangles have a hypotenuse that measures \(4 x+5\) inches and legs that measure \(3 x+13\) inches and \(x\) inches? What are the dimensions of the triangle(s)?

Challenge Problem Show that the real solutions of the equation \(a x^{2}+b x+c=0, a \neq 0,\) are the reciprocals of the real solutions of the equation \(c x^{2}+b x+a=0\), \(c \neq 0 .\) Assume that \(b^{2}-4 a c \geq 0\)

Geometric Mean If \(0
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