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faust18 [17]
3 years ago
9

Use foil to explain how to find the product of (a+b)(a-b). Then describe a shortcut that you could use to get this product witho

ut using foil.
Mathematics
2 answers:
chubhunter [2.5K]3 years ago
6 0
Using the FOIL method -
Firsts: a×a= a^2
Outers: a×-b= -ab
Inners: b×a= ab
Lasts: b×-b= -b^2

Combine -
a^2-ab+ab-b^2
= a^2-b^2

This particular pair of binomial brackets with a + and a - in them can be expanded with the difference of two squares:
(a+b)(a-b) = a^2-b^2
This works for binomial brackets with a + and a - in each of them, and where the two terms in each bracket are the sameFor example:
(x+5)(x-5)= x^2-25
So remember,
Difference
Of
Two
Squares,
Abbreviated as D.O.T.S
galben [10]3 years ago
4 0
FOIL stands for Front Outter Inner Last

F- multiply the first two terms axa= a^2
O- multiply the two terms on the outside a x -b= -ab
I- multiply the two inner (middle) terms - b x a= ab
L- multiply the last variable in each term= b^2


A shortcut without using FOIL is called the difference of squares and can be rewrote as (a^2-b^2)
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For this case we must solve the following quadratic equation:

3x ^ 2-2x + 5 = 0

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The roots are given by:

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Substituting the values we have:

x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {2 \pm \sqrt {4-60}} {6}\\x = \frac {2 \pm \sqrt {-56}} {6}

By definition we have to:

i ^ 2 = -1\\x = \frac {2 \pm \sqrt {56i ^ 2}} {6}\\x = \frac {2 \pm i \sqrt {56}} {6}\\x = \frac {2 \pm i \sqrt {2 ^ 2 * 14}} {6}\\x = \frac {2 \pm 2i \sqrt {14}} {6}\\x = \frac {1 \pm i \sqrt {14}} {3}

We have two complex roots:

x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}

Answer:

x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}

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