Problem 6 Concept Check Study this incorre... [FREE SOLUTION]

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Chapter 11: Problem 6

Concept Check Study this incorrect "solution." WHAT WENT WRONG? $$ \begin{aligned} 2(x-1)^{2}-3(x-1)+1 &=0 \\ 2 u^{2}-3 u+1 &=0 \\ & \text { Let } u=x-1 \\ (2 u-1)(u-1) &=0 \\ 2 u-1=0 & \text { or } u-1=0 \\ u=\frac{1}{2} & \text { or } \quad u=1 \end{aligned} $$ Solution set: $\left\\{\frac{1}{2}, 1\right\\}$

Short Answer

Expert verified

The correct solution set is {3/2, 2} after back-substitution.

Step by step solution

Identify the substitution

The variable substitution is identified correctly. Let u = x - 1.

Substitute into the equation

Substitute u = x - 1 into the original equation to get 2u^2 - 3u + 1 = 0.

Factor the quadratic equation

The factorization of the quadratic equation 2u^2 - 3u + 1 was performed correctly: (2u - 1)(u - 1) = 0.

Solve for u

Setting each factor equal to zero gives the solutions u = 1/2 or u = 1.

Back-substitute the variable

Substitute back x - 1 for u: x - 1 = 1/2 or x - 1 = 1.

Solve for x

Solve for x in each equation: x = 1 + 1/2 or x = 1 + 1. Hence, the solutions are x = 3/2 and x = 2.

Correct Solution Set

The correct solution set after solving for x is {3/2, 2}. This shows that the initial solution set {1/2, 1} was incorrect.

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Key Concepts

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

Quadratic Equations

Quadratic equations are polynomial equations of the form \[ ax^2 + bx + c = 0, \] where $ a, b,$ and $ c $ are constants, and $ a eq 0 $. Quadratic equations have important features such as:

They can be solved using different methods: factoring, completing the square, and the quadratic formula.
They often have two solutions, due to the nature of their graph, a parabola, intersecting the $ x $-axis at up to two points.

For example, in the given problem, the equation \[ 2(x - 1)^2 - 3(x - 1) + 1 = 0 \] needs to be solved. To do so, various strategies like variable substitution and factoring can be useful tools for finding the roots efficiently.

Variable Substitution

Variable substitution is a handy technique to simplify complex equations. In this approach, a new variable is introduced to replace an expression. This can significantly streamline the solving process.
In our exercise, we let \[ u = x - 1 \] to transform the original equation into a simpler form. This turns our complex quadratic into:
\[ 2u^2 - 3u + 1 = 0 \]
This substitution makes the equation easier to tackle using standard solving procedures for quadratics.

The benefit here is to change a shifted quadratic into a standard one.
After solving for the new variable $ u $, we will simply revert back to the original variable $ x $ by back-substituting.

Factoring

Factoring is a key technique for solving quadratic equations effectively. It involves expressing the quadratic equation as a product of two binomials. Here's the step-by-step factoring process used for our equation:
1. Begin with the simplified equation: $2u^2 - 3u + 1 = 0$.
2. Factor this into two binomials: $(2u - 1)(u - 1) = 0$.

Setting each factor to zero provides the solutions for $ u $:

\[ 2u - 1 = 0 \] \[ u - 1 = 0 \]

Solving these gives $ u = \frac{1}{2} $ or $ u = 1 $.

Finally, back-substitute $ u = x - 1 $ and solve for $ x $ to find the final solutions $ x = \frac{3}{2} $ and $ x = 2 $. Factoring transforms the quadratic into a solvable form, making the overall process more manageable and clear.

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Short Answer

Step by step solution

Identify the substitution

Substitute into the equation

Factor the quadratic equation

Solve for u

Back-substitute the variable

Solve for x

Correct Solution Set

Key Concepts

Quadratic Equations

Variable Substitution

Factoring

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