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vodomira [7]
3 years ago
7

How do you solve basic quadratic equations, using the "completing the square" method. I know how to use the quadratic formula bu

t don't know how to solve by completing the square.
Mathematics
1 answer:
Nitella [24]3 years ago
4 0
So, we have 80 = x² -16x.

first off, we start by grouping the terms with "x".

80 = (x² - 16x)

80 = (x² - 16x + [?]²)   <--- we have a missing fellow.

so, the idea being, we need "some value" to get our perfect square trinomial, hmmm what could that be?

well, the tell-tale guy is the middle term, we know that 2(√ guy on the left)(√ guy on the right) is the middle term.

we know the square root of the (√x²) is just "x", and the guy on the right is [?], but we also know that if we multiply 2 times both we get 16x, so

\bf 2(x)\left(\boxed{?}\right)=16x\implies \boxed{?}=\cfrac{16x}{2x}\implies \boxed{?}=8

aha!  there's our missing fellow, so we need to add 8².

however, bear in mind, all we're doing is borrowing from our very good friend Mr Zero, 0.

so if we add 8², we also have to subtract 8².

\bf 80=(x^2-16x+8^2-8^2)\implies 80=(x^2-16x+8^2)-64&#10;\\\\\\&#10;144=(x^2-16x+64)\implies 144=(x-8)^2\implies \pm\sqrt{144}=x-8&#10;\\\\\\&#10;\pm 12 = x-8\implies \pm 12 + 8 = x\implies x=&#10;\begin{cases}&#10;20\\&#10;-4&#10;\end{cases}
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Is -660.3 an integer?
Andreas93 [3]

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no

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Thanks for help have nice day
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7. b

8. d

9. c

Step-by-step explanation:

8 0
2 years ago
Plz i need help and plz explain
Montano1993 [528]

\frac{ {x}^{2}  - 7x + 12}{ {x}^{2} - x - 12 }  \\  =  \frac{ {x}^{2} - 4x - 3x + 12 }{ {x}^{2} - 4x + 3x - 12 }  \\  =  \frac{x(x - 4) - 3(x - 4)}{x(x - 4) + 3(x - 4)}  \\  =  \frac{(x - 3)(x - 4)}{(x + 3)(x - 4)}  \\  =  \frac{x - 3}{x + 3}

x² - x - 12 ≠ 0

x² - 4x + 3x - 12 ≠ 0

x (x - 4) + 3 (x - 4) ≠ 0

(x + 3)(x - 4) ≠ 0

x ≠ -3 or x ≠ 4

(B)

3 0
3 years ago
In each of problems 5 through 11, find the general solution of the given differential equation
Brrunno [24]

The complete question is

"Find the general solution of the given differential equation

y''-y=0, y1(t)=e^t , y2(t)=cosht

The function y(t)=e^t  is the solution of the given differential equation.

The function y(t)=cosht is the solution of given differential equation.

<h3>What is a function?</h3>

The function is a type of relation, or rule, that maps one input to specific single output.

Given;

y_1(t) = e^t

Given differential equations are,

y''-y = 0

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y' (t) = e^t, y'' (t) = e^t

Substitute values in the given differential equation.

e^t -e^t=0

                   

Therefore, the function y(t)=e^t  is the solution of the given differential equation.

Another function;

y(t)=cosht  

So that,  

y"(t)=sinht\\\\y"(t)=cosht

Hence, function y(t)=cosht is solution of given differential equation.

Learn more about function here:

brainly.com/question/2253924

#SPJ1

7 0
2 years ago
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