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Zepler [3.9K]
3 years ago
10

How would you find the solutions to this quadratic equation y=2x^2-6x-8

Mathematics
2 answers:
aleksandrvk [35]3 years ago
8 0

Answer:

Factoring.

General Formulas and Concepts:

<u>Algebra I</u>

  • Factoring
  • Finding roots of quadratics

Step-by-step explanation:

<u>Step 1: Define</u>

y = 2x² - 6x - 8

0 = 2x² - 6x - 8

<u>Step 2: Find roots</u>

  1. Factor 2:                    0 = 2(x² - 3x - 4)
  2. Factor:                        0 = 2(x - 4)(x + 1)
  3. Find roots:                 x = -1, 4

And we have our final answer!

Marysya12 [62]3 years ago
7 0

Answer:

first take 2 out of the equation

y=2(x^2-3x-4)

then factor

y=2(x-4)(x+1)

then solve

(x-4)=0 .  (x+1)=0

x=4 .          x=-1

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Answer:

y=1/4x-1.5

Step-by-step explanation:

So, because the line is parallel, the slope is the same. Then, to get the y-int, you do b=y-mx with the point, so the equation is b=-1-(1/4)(2), then you multiply to get b=-1-(0.5), so your y-int is -1.5.

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Help would be truly appreciated. Write the polynomial in standard form from the given zeroes of lest degree that has rational co
pishuonlain [190]

Answer:

1) The polynomial in standard form is f(x) = x³ - 9·x² + 23·x - 15

2) The polynomial in standard form is f(x) = x³ - (2 - 2·i)·x² + 4·i·x

3) The polynomial in standard form is f(x) = x² + (3·i - 3)·x + 2 - 6·i

4) The polynomial in standard form is f(x) = x² - (5 + √5)·x + 6 + 3·√5

Step-by-step explanation:

Given that that the polynomial is of least degree, we have;

The standard form is the form, f(x) = a·xⁿ + b·xⁿ⁻¹ +...+ c

1) The zeros of the polynomial are x = 5, 3, 1

Which gives the polynomial in factored form as_f(x) = (x - 5)·(x - 3)·(x - 1)

From which we have;

f(x) = (x - 5)·(x - 3)·(x - 1) = (x - 5)·(x² - 4·x + 3) = x³ - 4·x² + 3·x - 5·x² + 20·x - 15

f(x) = x³ - 9·x² + 23·x - 15

The polynomial in standard form is therefore f(x) = x³ - 9·x² + 23·x - 15

2) The zeros of the polynomial are x = 2, 0, 2·i

Which gives the polynomial in factored form as_f(x) = (x - 2)·(x)·(x - 2·i)

From which we have;

f(x) = (x - 2)·x·(x - 2·i) = (x² - 2·x)·(x - 2·i) = x³ - 2·i·x² + 2·x² + 4·ix

The polynomial in standard form is therefore f(x) = x³ - (2 - 2·i)·x² + 4·i·x

3) The zeros of the polynomial are x = 2, 1 - 3·i

Which gives the polynomial in factored form as_f(x) = (x - 2)·(x - 1 - 3·i)

From which we have;

f(x) = (x - 2)·(x - (1 - 3·i)) = x² - x + 3·i·x - 2·x + 2 -6·i = x² + 3·i·x - 3·x + 2 - 6·i

The polynomial in standard form is therefore f(x) = x² + (3·i - 3)·x + 2 - 6·i

4) The zeros of the polynomial are x = 3, 2 + √5

Which gives the polynomial in factored form as_f(x) = (x - 3)·(x - (2 + √5))

From which we have;

f(x) = (x - 3)·(x - (2 + √5)) = x² - 2·x - x·√5 - 3·x + 6 + 3·√5

f(x) = x² - 5·x - x·√5 + 6 + 3·√5 = x² - (5 + √5)·x + 6 + 3·√5

The polynomial in standard form is therefore f(x) = x² - (5 + √5)·x + 6 + 3·√5.

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Helga [31]

Answer: 689 + 30/32

Step-by-step explanation:

A reduction of A by B, means that we need to solve:

A - B.

So if we want to reduce 738 and 7/32 by 48 and 9/32 we have to solve:

(738 + 7/32) - (48 + 9/32)

we can separate it into whole and fraction:

(738 - 48) + (7/32 - 9/32)

690 - 2/32

But we usually don't want a negative fraction, so we can use that:

1 = 32/32

690 - 2/32 = 689 + 1 - 2/32 = 689 +32/32 - 2/32 = 689 + 30/32

8 0
3 years ago
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