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pychu [463]
3 years ago
15

Solve x 2 + 9x + 8 = 0 by completing the square. What are the solutions?

Mathematics
1 answer:
Sindrei [870]3 years ago
4 0

So firstly, what two terms have a product of 8x^2 and a sum of 9x? That would be x and 8x. Replace 9x with x + 8x: x^2+x+8x+8=0

Next, factor x^2 + x and 8x + 8 separately. Make sure that they have the same quantity on the inside: x(x+1)+8(x+1)=0

Now you can rewrite the equation as (x+8)(x+1)=0

Now using zero product property, solve for x:

x+8=0\\ x=-8\\ \\ x+1=0\\ x=-1

<u>In short, the solutions are {-1,-8}, or C.</u>

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8 0
3 years ago
If you are reading this please help!!!!
Soloha48 [4]

Given:

1.  -3 x^{2}\left(4 x^{3}-7\right)\\$2. $(6 x-5)(2 x+3)$\\3. $(3 x-1)\left(x^{2}+5 x-2\right)$

To find:

The product of the polynomials.

Solution:

1. -3 x^{2}\left(4 x^{3}-7\right) =-3 x^{2}(4 x^{3}) -3 x^{2}(-7)

Multiply the numerical coefficient and add the powers of x.

                          =-12 x^{5}+21 x^{2}

2. (6 x-5)(2 x+3)=6 x(2 x+3)-5(2x+3)

Multiply each term of first polynomial with each term of 2nd polynomial.

Multiply the numerical coefficient and add the powers of x.

                              =12 x^2+18x-10 x-15

                              =12 x^2+8x-15

3. (3 x-1)\left(x^{2}+5 x-2\right)=3 x(x^{2}+5 x-2)-1(x^{2}+5 x-2)

Multiply each term of first polynomial with each term of 2nd polynomial.

Multiply the numerical coefficient and add the powers of x.

                                      =3x^{3}+15 x^2-6x-x^{2}-5 x+2

Add or subtract like terms together.

                                      =3x^{3}+14 x^2-11x+2

The answer for multiplying polynomials:

1. -12 x^{5}+21 x^{2}

2. \  12 x^2+8x-15

3. \ 3x^{3}+14 x^2-11x+2

3 0
3 years ago
What is the value of x?
kolezko [41]
What is the question im solving x for?

6 0
3 years ago
Read 2 more answers
Can someone help me please. thanks!​
Luda [366]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Four times a number added to 3 times a larger number is 31. Seven subtracted from the larger number is equal to twice the smalle
aev [14]
Let the smaller number be x.
Let the bigger number be y.

\begin{cases} &4x + 3y = 31 \tex{ ----- (1) } \\ &y- 7 = 2x \tex{ ----- (2) } \end{cases}

Rearrange equation (2):
\begin{cases} &4x + 3y = 31 \tex{ ----- (1) } \\ &y = 2x + 7 \tex{ ----- (2) } \end{cases}&#10;


Sub (2) into (1):

4x + 3(2x + 7 ) = 31
4x + 6x + 21 = 31
10x + 21 = 31
10x = 10
x = 1

Sub x = 1 into (2):

y = 2x + 7
y = 2(1) + 7
y= 9

Answer: The two numbers are 1 and 9.

5 0
3 years ago
Read 2 more answers
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