A student was asked to solve the equation `2x^2+4x-4=0` the students incorrect attempt is shown below.` ` `2x^2+4x-4=0` `x^2+2x-2=0` `(x+2)^2-6=0` `(x+2)^2=6` `(x+2)=sqrt(6)` `=-2+sqrt(6)`...

A student was asked to solve the equation `2x^2+4x-4=0`

the students incorrect attempt is shown below.` `

`2x^2+4x-4=0`

`x^2+2x-2=0`

`(x+2)^2-6=0`

`(x+2)^2=6`

`(x+2)=sqrt(6)`

`=-2+sqrt(6)`

identify the two mistakes made by the student explain as if directly to the student why their workings is incorrect

Asked on by dani1985

4 Answers | Add Yours

steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

The factoring in line 3 is incorrect.  Here, they used the "completing the square" method.  There can be tweeks to it, but I show it like this.

After line 2, identify a, b, and c:

a = 1, b = 2, c = -2

Take half of b and square that result:

b/2 = 1, 1^2 = 1

Add that to each side, but don't combine it to anything on the left, as such:

x^2 + 2x + 1 - 2 = 0 + 1

Then, the first 3 terms on the right side can factor

(x^2 + 2x + 1) - 2 = 1

(x+1)^2 - 2 = 1

From here, the "method" they used was correct except at the second to last step.  So, you would move the number outside the parenthesis over

(x+1)^2 - 2 + 2 = 1 + 2

(x+1)^2 = 3

Then, square root each side.  This is their second error.  When you square root the right side, you would get the positive and negative square roots of the number.

x+1 = +sqrt(3)

Then, subtract 1:

x = -1 + sqrt(3)

rimmery's profile pic

rimmery | Student, Undergraduate | (Level 1) Honors

Posted on

The first mistake is between the 2nd and 3rd line, when you try to factor `x^2+2x-2`

The factoring is wrong in that `2x` is not part of `(x+2)^2` and therefore it doesn't matter what number you add to try and complete the square with, it won't factor.

The second mistake is in between the 4th and 5th line, when you square root both sides. Remember that the square of a negative number is always a positive number, so  `sqrt(6)^2=6 and (-sqrt(6))^2=6`

Now, since the factoring is wrong, this entire equation is wrong, so going back to line 2 and completing the square:

To complete the square for this equation, we need to try to get it in the form of `(x+1)^2`

 and not `(x+2)^2`  as the student has done originally, because the `2x` cannot be factored as part of `(x+2)^2`  

So to correctly complete the square, you would add/subtract 3 to the left hand side, so we have

`x^2+2x-2+3-3=0`  

`(x^2+2x+1)-3=0`

`(x+1)^2-3=0`

`(x+1)^2=3`

`(x+1)=sqrt(3) or (x+1)=-sqrt(3)`

`x=sqrt(3)-1 or x=-sqrt(3)-1`

` `

ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

Posted on

The first mistake can be seen in step 3. In order to factor, the student added -6 to the equation. However, you have to add -6 to both sides of the equation in order for it to be mathematically correct. The student should remember that whatever you do on one side, you have to do on the other.

Another mistake could be ignoring the possibility of the answer being a positive AND negative number in step 5. To be mathematically correct (ignoring the first mistake), the answer should be `-2 +` `sqrt(6)`  and `-2 -``sqrt(6)` .

dani1985's profile pic

dani1985 | Student, Kindergarten | (Level 1) eNoter

Posted on

edit to the above question:

identify the two mistakes made by the student. explain as if directly to the student, why their workings are incorrect.

We’ve answered 318,917 questions. We can answer yours, too.

Ask a question