Question

Freddy has been asked to solve the following expressions by factoring.Equation 1: x^2 + 8x + 12Equation 2: 2x^2 - 9x + 10Here are Freddy's solutions; both solutions are incorrect. Analyze both of Freddy'ssolutions and answer the questions below:Freddy's Equation 1x^2 + 8x + 12step 1: (x )(x )step 2: (x 3) (x 4) ‹- Freddy later scratched this outstep 3: (x 6) (x 2)step 4: (x + 6) (x + 2)step 5: (x + 6)(c + 2) =0step 6: 2 = 6 and r =2 Freddy's Equation 2:2x^2 - 9x + 10step 1: (x )(x )step 2: (x 10) (x 1)step 3: (x - 10) (x + 1)step 4: (x-10)（x +1)=0step 5: x = 10 and x= -1Analyze both of Freddy's solutions and do the following:a. Explain where Freddy's logic broke down.b. Explain your conclusions about Freddy's answers.c. Show how you would teach Freddy to factor these expressions to solve for x.d. Explain how to solve Equation 1 by completing the square.

Answer 1

Part a)

In the first factorization, Freddy made two mistakes. The first mistake is in step 2 (Which he deleted later) because those two numbers (3 and 4) do not complete the square. The second mistake is more subtle because he factorized correctly but he did not find the correct root.

When you have a factorization in the form of (x-a) (x-b) = 0, the roots (or solutions) can be found if either (x-a) = 0 or (x-b) = 0 :

`\begin{gathered} (x-a)(x-b)=0 \\ This\text{ is true only if:} \\ (x-a)=0\text{ or \lparen}x-b)=0 \\ \text{ Then the solutions are:} \\ x_1=a\text{ and }x_2=b \end{gathered}`

In this first factorization: ( x - (-6) ) ( x - (-2) ) = 0, a = -6 and b = -2.

In the second factorization, he made a mistake assigning the signs of the numbers 10 and 1; and separating the equation into two parts. If you want to factorize correctly a quadratic equation of the form ax^2+bx+c = 0, the first step is:

`\begin{gathered} (ax^2+bx+c)=0\text{ Multiply both sides for a and divide it by a} \\ ((ax)^2+b(ax)+ac))/(a)=0 \\ (ax+x_1)(ax+x_2)=0 \end{gathered}`

*Notice that the "plus signal" in the first factor comes from the sign of b, i.e. if b is negative then the first factor is negative. However, the second "plus signal" comes from the product of the signs that accompanies b and c.

Part b)

Both answers are wrong because he did not find the roots correctly in the first factorization and he did not factorize correctly the quadratic equation in the second equation.

Part c)

First equation: The solutions are -2 and -6.

`\begin{gathered} x^2+8x+12=0\text{ \lparen this has the form of x\textasciicircum2+bx+c=0\rparen} \\ (x+m)(x+n)=0\text{ \lparen The values of ''m'' and ''n'' must satisfy that m*n=c and m+n=b} \\ (x+6)(x+2)=0\text{ \lparen These are the values because 6*2= 12 and 6+2=8\rparen} \\ \text{\lparen If any of the factors is zero, then the whole equation is zero\rparen} \\ (x_2+2)=0\rightarrow x_2=-2 \\ (x_1+6)=0\rightarrow x_1=-6 \end{gathered}`

Second equation:

`\begin{gathered} 2x^2-9x+10=0\text{ \lparen This is an equation of the form ax\textasciicircum2+bx+c=0\rparen} \\ ((2x)^2-9(2x)+20)/(2)=0\text{ \lparen Now the equation has the form of x\textasciicircum2+bx+x=0, just that x = 2m} \\ ((2x-m)(2x-n))/(2)=0\text{ \lparen We have to find m and n, that mn=c, and m+n=b} \\ ((2x-5)(2x-4))/(2)=0\text{ \lparen This is true because \lparen5*4\rparen=20 and 5+4=9} \\ (2x-5)(x-2)=0\text{ \lparen We divided by two the last part\rparen} \\ x_1=(5)/(2) \\ x_2=2 \end{gathered}`

Part d)

To solve equation 1 using the technique of completing the square is just how I solved it before. You first separate them into two factors, then you find the signs of those factors (the first is the sign of b, and the second is the sign of b*c), and finally, you have to find the values of m and n, that satisfy the condition of m+n = b and mn=c.