/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Solve the quadratic equation usi... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the quadratic equation using any method. Find only real solutions. $$x^{2}-2 x=9$$

Short Answer

Expert verified
The real solutions to the equation are \( x = 3 \) and \( x = -3 \).

Step by step solution

01

Rearrange the equation

Rearrange the equation into the standard quadratic form \( ax^2 + bx + c = 0 \). \n To do this, subtract 9 from both sides. The equation now becomes \( x^2 - 2x - 9 = 0 \).
02

Factorise the quadratic

Now that the equation is in the standard quadratic form, look for two numbers that add to -2 (the coefficient of x) and multiply to -9 (the constant term). The numbers -3 and 3 fit these criteria.\n So, the quadratic can be factorised to \( (x - 3)(x + 3) = 0 \).
03

Solve for x

A product of two factors equals zero if and only if at least one of the factors is zero. So set each factor to zero and solve for x: \n If \( x - 3 = 0 \), then \( x = 3 \) \n If \( x + 3 = 0 \), then \( x = -3 \).

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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 method used to solve quadratic equations, which are equations of the second degree usually in the form of \( ax^2 + bx + c = 0 \). The goal of factoring is to break down the quadratic expression into the product of simpler binomials. Imagine a quadratic equation as a trinomial that you want to rewrite as \( (x + m)(x + n) = 0 \), where \( m \) and \( n \) are the numbers that when multiplied give you the constant term, \( c \) and when added, give you the linear coefficient, \( b \) of the quadratic equation.

To factor our given equation \( x^2 - 2x - 9 = 0 \), identify two numbers that multiply to -9 and add up to -2—these are -3 and 3. Hence, we get two factors, \( x - 3 \) and \( x + 3 \) which are set to zero to find the values of \( x \) that satisfy the equation. This approach simplifies the solving process and helps reveal the real solutions efficiently.
Quadratic Formula
When factoring is challenging or if the quadratic cannot be factored easily, the quadratic formula provides an alternative and uniform method to find the solutions. It is derived from completing the square on the general quadratic equation \( ax^2 + bx + c = 0 \). The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). This formula allows us to calculate the solutions of any quadratic equation by substituting the coefficients \( a \), \( b \), and \( c \) into the formula.

The discriminant, \( b^2-4ac \), determines the nature of the roots. If the discriminant is positive, there are two distinct real solutions. If it's zero, there's exactly one real solution. A negative discriminant indicates that no real solutions exist. Thus, the quadratic formula not only gives the specific values for the solutions but also informs us about the nature of the solutions based on the discriminant.
Real Solutions of Quadratics
Real solutions of quadratics are the x-values that satisfy the equation \( ax^2 + bx + c = 0 \) where all coefficients are real numbers, and the solutions are also real numbers. These solutions can be visualized as the points where the graph of the quadratic equation crosses the x-axis. The discriminant \( b^2-4ac \) is key in determining the number of real solutions. A discriminant greater than zero means two real solutions, equal to zero indicates one real solution (a repeated root), and less than zero implies there are no real solutions, only complex solutions.

For the example \( x^2 - 2x - 9 = 0 \) after factoring yields real solutions \( x = 3 \) and \( x = -3 \) by setting \( x - 3 \) and \( x + 3 \) to zero, respectively. These solutions can also be verified using the quadratic formula, which should produce the same real solutions, ensuring the consistency of both methods.
Standard Quadratic Form
The standard quadratic form is the format of a quadratic equation expressed as \( ax^2 + bx + c = 0 \) with \( a \) not equal to zero. Recognizing this form is critical because it is the starting point for finding solutions through factoring, completing the square, or using the quadratic formula.

In the initial exercise, we transformed \( x^2 - 2x = 9 \) into the standard form by subtracting 9 from both sides, resulting in \( x^2 - 2x - 9 = 0 \). This maneuver sets the stage for applying strategies like factoring. When teaching or solving quadratic equations, always ensure that the equation is in the standard form first. This step is a fundamental part of the process and aids students in approaching the problem systematically and using various methods to find the solutions.

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Most popular questions from this chapter

When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=x-2 ; g(x)=2 x^{2}-x+3$$

Graph each quadratic function by finding a suitable viewing window with the help of the TABLE feature of a graphing utility. Also find the vertex of the associated parabola using the graphing utility. $$s(t)=-16 t^{2}+40 t+120$$

Graph the function \(f(t)=t^{2}-4\) in a decimal window. Using your graph, determine the values of \(t\) for which \(f(t) \geq 0\).

Let \(g(s)=-2 s^{2}+b s .\) Find the value of \(b\) such that the vertex of the parabola associated with this function is (1,2)

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