Marla made an error when solving the equation shown. What was her error ,and how should she fix it?
1 answer:
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Answer:
<em><u>t</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em>
<em><u>hope</u></em><em><u> </u></em><em><u>it help you</u></em>
<em><u>thanks</u></em><em><u>!</u></em>
You can get all the terms on one side, and then solve for x by any means:
Factor out a -2:
Factor this equation:
I will use the AC method. To use it, first multiply a and c (in ax^2 + bx +c):
Now, look for two numbers that multiply to -12 and add to -11. Obviously these numbers are -12 and 1. (-12*1 = -12 and -12+1 = -11). Now, because of rules, you set it up in a Punnett square:
4x^2 -12x
x -3
Now, we find common factors of the terms in rows:
_x_______-3__
4x| 4x^2 -12x
1 | x -3
So, you can use this to write an equivalent expression to the quadratic given:
Now, we known the factors are (solve for x): 3 and -1/4.
Factor out a -2:
Factor this equation:
I will use the AC method. To use it, first multiply a and c (in ax^2 + bx +c):
Now, look for two numbers that multiply to -12 and add to -11. Obviously these numbers are -12 and 1. (-12*1 = -12 and -12+1 = -11). Now, because of rules, you set it up in a Punnett square:
4x^2 -12x
x -3
Now, we find common factors of the terms in rows:
_x_______-3__
4x| 4x^2 -12x
1 | x -3
So, you can use this to write an equivalent expression to the quadratic given:
Now, we known the factors are (solve for x): 3 and -1/4.
