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nignag [31]
3 years ago
11

Compare and Contrast how you would factor x^2−9 and x^2+9. For full credit, make sure you explain why they are factored differen

tly.
Mathematics
1 answer:
marishachu [46]3 years ago
7 0

The factored form of the expression x^2 - 9 is (x+3)(x-3)

The expression x^2 + 9 has no real roots hence cannot be factored

Quadratic equations are equations with a degree of 2.

Given the quadratic function x^2-9 \ and \ x^2+9

For the function x^2 - 9

=x^2-9\\=x^2-3^2

Using the difference of two squares:

a^2-b^2=(a-b)(a+b)\\x^2-3^2=(x-3)(x+3)

For the expression x^2 + 9

The expression x^2 + 9 has no real roots hence cannot be factored

Learn more on factorization here: brainly.com/question/20293447

You might be interested in
Find an m > 0 such that the the equation x^4−(3m+2)x^2+m^2=0 has four real solutions that form an arithmetic sequence.
Aleonysh [2.5K]

Answer:

The value of m is 6.

Step-by-step explanation:

Here, the given equation,

x^4-(3m+2)x^2+m^2=0

x^4+0x^3-(3m+2)x^2+0x+m^2=0

Let the roots of the equation are a-3b, a-b, a+b and a + 3b, ( they must be form an AP )

Thus, we can write,

a-3b+a-b+a+b+a+3b=\frac{\text{coefficient of }x^3}{\text{coefficient of }x^4}

=\frac{0}{1}=0

\implies a=0----(1)

(-3b)(-b)+(-b)(b)+(b)(3b)+(3b)(-3b)+(-b)(3b)+(-3b)(b)=\frac{\text{coefficient of }x^2}{\text{coefficient of }x^4}}

=\frac{-3m-2}{1}

3b^2-b^2+3b^2-9b^2-3b^2-3b^2=-3m-2

-10b^2=-3m-2

\implies b^2=\frac{3m+2}{10}-----(2)

(-3b)(-b)(b)(3b)=\frac{\text{Constant term}}{\text{coefficient of}x^4}= m^2

9b^4=m^2

9(\frac{3m+2}{10})^2=m^2

9(\frac{9m^2+4+12m}{100})=m^2

81m^2+36+108m=100m^2

-19m^2+108m+36=0

19m^2-108m-36=0

19m^2-114m+6m-36=0

19m(m-6)+6(m-6)=0

(19m+6)(m-6)=0

\implies m=-\frac{6}{19}\text{ or }m=6

But m > 0,

Hence, the value of m is 6.

4 0
3 years ago
Let log_(b)A=3; log_(b)C=2; log_(b)D=5 what is the value of log_(b)((A^(5)C^(2))/(D^(6)))
Colt1911 [192]

Answer:

-11

Step-by-step explanation:

Given

\log_bA=3\\ \\\log_bC=2\\ \\\log_bD=5

Use properties:

\log_b(A\cdot C)=\log_bA+\log_bC\\ \\\log_b\dfrac{A}{C}=\log_bA-\log_bC\\ \\\log_bA^k=k\log_bA

Thus,

\log_b\dfrac{A^5\cdot C^2}{D^6}\\ \\=\log_b(A^5\cdot C^2)-\log_bD^6\\ \\=\log_bA^5+\log_bC^2-\log_bD^6\\ \\=5\log_bA+2\log_bC-6\log_bD\\ \\=5\cdot 3+2\cdot 2-6\cdot 5\\ \\=15+4-30\\ \\=-11

3 0
3 years ago
F(x+k) what is the value of k
guapka [62]

Answer:

k = -3

Hope this helps!

6 0
3 years ago
Read 2 more answers
HELP SOMEONE ASAP LOTS OF POINTS
PSYCHO15rus [73]

Answer:

3 3/8

Step-by-step explanation:

4          5           27

---   -   -----   =  -------  or 3 3/8

1           8           8

7 0
3 years ago
What is the solution to the linear equation d-10-2d+7=8+d-10-3d
bija089 [108]
D-10-2d+7 = 8 + d- 10 -3d
d-2d-10+7 = 8-10 +d -3d
-d-3=-2-2d
-d+2d=3-2
d=1

Final answer: C. d=1
8 0
3 years ago
Read 2 more answers
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