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Chapter 6: Problem 68

Solve each equation. See Examples I through 9. (A few exercises are linear equations.) $$ x^{2}+22 x+121=0 $$

Short Answer

Expert verified
The solution to the equation is \( x = -11 \).

Step by step solution

01

Identify the Equation Type

The given equation is a quadratic equation in the standard form, where the general structure is \( ax^2 + bx + c = 0 \). In this case, \( a = 1 \), \( b = 22 \), and \( c = 121 \).
02

Recognize Perfect Square

Notice that the constant term \( c = 121 \) is a perfect square, specifically \( 11^2 \), and the middle term \( 22x \) is twice the product of 11 and x, which indicates the equation might be a perfect square trinomial.
03

Rewrite the Equation

Rewrite the quadratic equation in the form of a perfect square: \((x + 11)^2 = 0\). This is done because \( (x+11)(x+11) = x^2 + 22x + 121 \).
04

Solve for x

To find the value of \( x \), take the square root of both sides of the equation \( (x+11)^2 = 0 \), resulting in \( x + 11 = 0 \).
05

Solve the Linear Equation

Subtract 11 from both sides to solve for \( x \): \( x = -11 \).

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Key Concepts

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

Quadratic Equation
A quadratic equation is a type of polynomial equation in which the highest power of the variable is 2. It's typically written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In our exercise, this form was \( x^{2} + 22x + 121 = 0 \). Quadratic equations can have different types of solutions, ranging from two real solutions, one real solution, or two complex solutions, depending on the discriminant鈥檚 value derived from the quadratic formula.
  • Two real and distinct solutions if \( b^2 - 4ac > 0 \).
  • One real and repeated solution if \( b^2 - 4ac = 0 \).
  • Two complex solutions if \( b^2 - 4ac < 0 \).
In this particular case, the equation factors into a perfect square, leading to one real and repeated solution.
Perfect Square Trinomial
A perfect square trinomial is a special type of quadratic equation that can be expressed as the square of a binomial. It takes the general form of \((x + m)^2 = x^2 + 2mx + m^2\) or \((x - m)^2 = x^2 - 2mx + m^2\). In our original problem, we encountered the trinomial \( x^2 + 22x + 121 \), which can be rewritten as the square of a binomial: \((x + 11)^2\).
  • The first term \( x^2 \) comes from \( x \times x \).
  • The middle term \( 22x \) results from \( 2 \cdot 11 \cdot x \), i.e., twice the product of \( 11 \) and \( x \).
  • The last term \( 121 \) is \( 11^2 \), confirming it is a perfect square.
Recognizing perfect square trinomials can simplify solving quadratic equations, as you can directly take the square root to find solutions.
Linear Equations
Once a quadratic equation is rewritten as a perfect square trinomial, it simplifies to a form where we solve it much like a linear equation. In simpler terms, a linear equation is an equation that makes a straight line when it is graphed. Its standard form is \( ax + b = 0 \), with solutions easy to find by isolating \( x \). In our exercise, after transforming the quadratic into \( (x + 11)^2 = 0 \), we derived \( x + 11 = 0 \), a linear equation. Solving this involves basic algebraic steps:
  • Subtract \( 11 \) from both sides to isolate \( x \).
  • The solution \( x = -11 \) gives the point where the equation is satisfied.
This reduction step is crucial as it transforms complex quadratic forms into simpler linear ones easily solvable by typical algebraic manipulation.

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Most popular questions from this chapter

Factor each trinomial. (Hint: Notice that \(x^{2 n}+4 x^{n}+3\) factors as \(\left.\left(x^{n}+1\right)\left(x^{n}+3\right) . \text { Remember } x^{n} \cdot x^{n}=x^{n+n} \text { or } x^{2 n} .\right)\) $$ x^{2 n}+8 x^{n}-20 $$

Solve each equation. See Examples I and 2. $$ x(x-7)=0 $$

An object is thrown upward from the top of an 80 -foot building with an initial velocity of 64 feet per second. The height of the object after \(t\) seconds is given by \(-16 t^{2}+64 t+80\). Factor this polynomial.

Complete each sentence in your own words. If \(x^{2}+b x+c\) is factorable and \(c\) is positive, then the signs of the last-term factors of the binomials are the same because....

Factor out the GCF from each polynomial. See Examples 4 through 10. $$ 6 x^{3}-9 x^{2}+12 x $$
See all solutions

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