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BARSIC [14]
3 years ago
11

Find the real solutions of the equation. (2x-6)^2 + 4(2x -6) +3 = 0

Mathematics
1 answer:
miskamm [114]3 years ago
7 0

Answer:

(

2

x

−

6

)

2

+

4

(

2

x

−

6

)

+

3

=

0

Simplify the left side.

Tap for more steps...

(

2

x

−

6

)

2

+

8

x

−

21

=

0

Use the quadratic formula to find the solutions.

−

b

±

√

b

2

−

4

(

a

c

)

2

a

Substitute the values  

a

=

4

,  

b

=

−

16

, and  

c

=

15

into the quadratic formula and solve for  

x

.

16

±

√

(

−

16

)

2

−

4

⋅

(

4

⋅

15

)

2

⋅

4

Simplify.

Tap for more steps...

x

=

4

±

1

2

The final answer is the combination of both solutions.

x

=

5

2

,

3

2

Step-by-step explanation:

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shepuryov [24]

Answer:

3

Step-by-step explanation:

=12/4

the common ratio =3

3 0
3 years ago
Drag the expressions into the boxes to correctly complete the table.
lora16 [44]

Answer:

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form ax^n.

Here:

n = non-negative integer

a = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

x^4+\frac{5}{x^3}-\sqrt{x}+8

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains \sqrt{x}, so it is not a polynomial.

Also it contains the term \frac{5}{x^3} which can be written as 5x^{-3}, meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression x^4+\frac{5}{x^3}-\sqrt{x}+8 is not a polynomial.

Given the expression

-x^5+7x-\frac{1}{2}x^2+9

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

\left|x\right|^2+4\sqrt{x}-2

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

x^3-4x-3

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

\frac{4}{x^2-4x+3}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

3 0
3 years ago
PLZ HELP I PROMISE I WILL GIVE BRAINLIEST!!!! ASAP
irga5000 [103]

Answer:

4c^{7}d^{13}

Step-by-step explanation:

(2cd^{4} )^{2} *(cd)^{5}

4c^{2}d^{8} *c^{5} d^{5}

4c^{7}d^{13}

4 0
3 years ago
Please help me on this <br><br> Simplify (12^2)^4
nika2105 [10]

(12^2)^4

= 12^(2*4)

= 12^8 (or 429,981,696)

7 0
3 years ago
Select all that apply.
dimulka [17.4K]
2x 2 - 6x = 0 .... if you mean 2x^2 - 6x = 0
then 2x = 0 and x - 3 = 0 << answer
3 0
3 years ago
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