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zlopas [31]
3 years ago
12

Factor 7(x − 3)2 − 4(x − 3) − 3 completely.

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
7 0
7(x-3)^2-4(x-3)-3\\\\substitute\ t=x-3,\ then\ we\ have\\\\7t^2-4t-3\\\\a=7;\ b=-4;\ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot7\cdot(-3)=16+84=100\\\\t_1=\dfrac{-b-\sqrt\Delta}{2a};\ t_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\t_1=\dfrac{-(-4)-10}{2\cdot7}=\dfrac{4-10}{14}=\dfrac{-6}{14}=-\dfrac{3}{7}\\\\t_2=\dfrac{-(-4)+10}{2\cdot7}=\dfrac{4+10}{14}=\dfrac{14}{14}=1\\\\7t^2-4t-3=7\left(t+\dfrac{3}{7}\right)(t-1)=(7t+3)(t-1)\\\\=(7(x-3)+3)(x-3-1)=(7x-21+3)(x-4)\\\\=\boxed{(7x-18)(x-4)}
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Answer: The system of equations are;

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a + 3b = 23———(2)

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From equation (1), what we have is the total number of shots he has taken altogether which is 9 shots in all. All 9 shots are an addition of free throws and three pointers (that is a + b).

In equation (2), what we have is the points obtainable times the number of shots taken (for each shot). This means if a is a free throw, then 1 times a is equal to number of free throws times 1. Similarly, if b is a three-point throw, then 3 times b is equal to the number of three pointers thrown times 3.

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4 years ago
Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Us
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ANSWER:

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