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3 years ago

What is a quadratic equation? And how do you solve them?

Mathematics

1 answer:

Allisa [31]3 years ago

3 0

Answer:

A quadratic equation is some thing like

(x+2)^2

(x+2)(x+5)

In order to solve them, the most basic method is the FOIL method

F-First

O-Outside

I-Inner

L-Last

So for example

(x+2)^2

Which is basically

(x+2)(x+2)

So the first is the 2 Xs

(x+2)(x+2)

x*x=x^2

The outer is the x and the 2,

(x+2)(x+2)

2*x=2x

Now the Inner which is the 2 and the x

(x+2)(x+2)

2*x=2x

And finally the last which is the 2 and the second 2

(x+2)(x+2)

2*2=4

Now add them all up

X^2+2x+2x+4

Combine like terms

x^2+4x+4

There you have it

Do the same with every other quadratic equation

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Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4

riadik2000 [5.3K]

<h3>The complete question:</h3>

Harold uses the binomial theorem to expand the binomial $(3x^5 -\dfrac{1}{9}y^3)^4$

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer:

(a). $(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$$

(b). $(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}$

Step-by-step explanation:

(a).

The binomial theorem says

$(x+y)^n=$$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $$$

For our binomial this gives

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $$}$

(b).

We simplify the terms of the expansion and get:

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4$

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4$

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561} }$

3 0

3 years ago

A magazine sales person earns $250 a week plus a commission of of the total sales. If the total sales for the week were $693, wh

Ira Lisetskai [31]

Answer:

Your answer would be C

Hope this helps

(\) QueTooOfficial (/)

4 0

3 years ago

I WILL GIVE YOU 40 POINTS PLEASE HELP (2)/(x^2-9) - (3x)/( x^2-5x+6) please show work

mars1129 [50]

Answer:

$\frac{-3x^2-7x-4}{x^3-2x^2-9x+18}$

Step-by-step explanation:

$\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factor x²-9 and x²-5x+6.

$\frac{2}{\left(x+3\right)\left(x-3\right)}-\frac{3x}{\left(x-2\right)\left(x-3\right)}$

Least common multiple of (x+3), (x-3), (x-2), and (x-3) is (x+3), (x-3), and (x-2).

Adjust the fractions based on LCM.

$\frac{2\left(x-2\right)}{\left(x+3\right)\left(x-3\right)\left(x-2\right)}-\frac{3x\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}$