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Find all real solutions. Do not use a calculator. $$12 x^{3}=17 x^{2}+5 x$$

Short Answer

Expert verified
The real solutions are \( x = 0 \), \( x = \frac{5}{3} \), and \( x = -\frac{1}{4} \).

Step by step solution

01

Rearrange the Equation

Start by moving all terms to one side of the equation to set it to zero: \[ 12x^3 - 17x^2 - 5x = 0 \] This gives us a standard form polynomial to solve.
02

Factor Out the Common Term

Notice that each term in the equation has a common factor of \( x \). Factor \( x \) out of the equation: \[ x(12x^2 - 17x - 5) = 0 \] This gives a linear factor \( x = 0 \) and a quadratic factor \( 12x^2 - 17x - 5 = 0 \).
03

Solve the Linear Factor

The factor \( x = 0 \) provides one solution. Thus, one real solution is:\[ x = 0 \]
04

Solve the Quadratic Factor

Use the quadratic formula to solve \( 12x^2 - 17x - 5 = 0 \). The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 12 \), \( b = -17 \), and \( c = -5 \).
05

Compute the Discriminant

Calculate the discriminant \( b^2 - 4ac \):\[ (-17)^2 - 4(12)(-5) = 289 + 240 = 529 \] As the discriminant is a perfect square, there are two distinct real solutions.
06

Calculate the Quadratic Solutions

Use the quadratic formula:\[ x = \frac{-(-17) \pm \sqrt{529}}{2(12)} \] \[ x = \frac{17 \pm 23}{24} \] This results in two solutions: \[ x = \frac{40}{24} = \frac{5}{3} \] \[ x = \frac{-6}{24} = -\frac{1}{4} \]
07

Compile All Solutions

Combining all solutions from steps 3 and 6, the real solutions are:1. \( x = 0 \) 2. \( x = \frac{5}{3} \) 3. \( x = -\frac{1}{4} \)

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Key Concepts

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

Factoring Polynomials
Factoring polynomials is like breaking large numbers into smaller parts, making them easier to manage and solve. It's an essential skill in algebra and helps simplify polynomial equations. Essentially, factoring involves finding expressions that multiply together to give the original polynomial. In this exercise, we began by recognizing that all terms in the given equation share a common factor. By identifying the greatest common factor (GCF), in this case, \( x \), we can simplify the equation significantly. Doing so reduces the degree of the polynomial and separates it into more manageable parts.
When dealing with polynomial equations, the goal is often to express the polynomial as a product of its factors. This technique turns difficult problems into simpler ones, typically resulting in a combination of linear and quadratic expressions.
  • Finding the GCF first makes subsequent factoring steps more manageable and straightforward.
  • After factoring, solving becomes easier because dealing with lower-degree polynomials is generally less complex.
This step-by-step factoring can be done manually with practice, setting the stage for more advanced techniques later on.
Quadratic Formula
The quadratic formula is a universal tool used when factoring a quadratic equation directly is too complex. It solves any quadratic equation in the form \( ax^2 + bx + c = 0 \). This formula draws its power from using the coefficients of the quadratic to find the solutions directly. For our exercise, once we factored out the common \( x \), we were left with a quadratic equation: \( 12x^2 - 17x - 5 = 0 \).
Applying the quadratic formula requires three main steps:
  • Identify the coefficients \( a \), \( b \), and \( c \).
  • Calculate the discriminant \( b^2 - 4ac \), which determines the nature of the roots.
  • Substitute the values into the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
The quadratic formula is especially useful because it provides solutions even when a polynomial doesn't easily factorize. Its precision and general applicability make it indispensable for solving second-degree polynomials.
Real Solutions
Real solutions refer to the values of \( x \) that satisfy a polynomial equation and are real numbers. In our problem, we aimed to find these solutions without a calculator, relying purely on algebraic methods. After factoring and using the quadratic formula, we discovered both simple and more complex solutions.
For the linear factor, \( x = 0 \) was straightforward, showing the importance of factoring common terms first.
The quadratic part required the discriminant calculation, a crucial step which reveals whether solutions are real, repeated, or complex:
  • A positive discriminant indicates two real and distinct roots. This was the case here with a value of 529, which is a perfect square, leading to nice integer solutions.
  • A zero discriminant would suggest a single, repeated real root.
  • A negative discriminant points to complex roots, outside the scope of this real solution exercise.
By combining solutions from both linear and quadratic parts, the full set of real solutions can be confidently identified. In our scenario, these were \( x = 0 \), \( x = \frac{5}{3} \), and \( x = -\frac{1}{4} \), demonstrating our comprehensive approach to solving polynomial equations.

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Most popular questions from this chapter

Use a graphing calculator to find a comprehensive graph and answer each of the following. (a) Determine the domain. (b) Determine all local minimum points, and tell if any is an absolute minimum point. (Approximate coordinates to the nearest hundredth.) (c) Determine all local maximum points, and tell if any is an absolute maximum point. (Approximate coordinates to the nearest hundredth.) (d) Determine the range. (If an approximation is necessary. give it to the nearest hundredth.) (e) Determine all intercepts. For each function, there is at least one \(x\) -intercept that has an integer x-value. For those that are not integers, give approximations to the nearest hundredth. Determine the \(y\) -intercept analytically. (f) Give the open interval(s) over which the function is increasing. (g) Give the open interval(s) over which the function is decreasing. $$P(x)=2 x^{4}+3 x^{3}-17 x^{2}-6 x-72$$

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=x^{3}+x^{2}-21 x-45 ; \quad k=-3$$

Use the given zero to completely factor \(P(x)\) into linear factors. Zero: \(1+i ; \quad P(x)=x^{4}-2 x^{3}+3 x^{2}-2 x+2\)

Use Descartes' rule of signs to determine the possible numbers of positive and negative real zeros for \(P(x) .\) Then use a graph to determine the actual numbers of positive and negative real zeros. $$P(x)=2 x^{5}-x^{4}+x^{3}-x^{2}+x+5$$

Answer true or false to each statement. Then support your answer by graphing. The graph of \(f(x)=x^{3}-3 x^{2}+3 x-1=(x-1)^{3}\) has exactly one \(x\) -intercept.

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