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svlad2 [7]
2 years ago
10

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

Mathematics
1 answer:
enyata [817]2 years ago
5 0

Answer:

-2,  (-5 is extraneous)

Step-by-step explanation:

assuming √(x+6) -4 =x

√(x+6) = x+4

square both side

x+6 = (x+4)^2

x+6 = x^2+8x+16

x^2+7x+10=0

(x+2)(x+5) = 0

x = -2, -5

put -5 back into org equation and it is not true

√(-5+6) -4 ≠ -5

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Factor completely 3x2 + 6x - 7x - 14
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Step-by-step explanation:

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The graph below represents the solution set of which inequality?
natulia [17]

Answer:

option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

⇒  x(x-4)+2(x-4)=0

⇒ (x+2)(x-4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x+2>0 and x-4

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x-2>0 and x+4

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

x+2 and x-4

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

⇒ x(x+4)-2(x+4)>0

⇒ (x-2)(X+4)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x-2 and x+4

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




5 0
3 years ago
Which expression uses the associative property to make it easier to evalute?
mixas84 [53]

Answer:

The correct option is (c).

Step-by-step explanation:

The given expression is :

14(\dfrac{3}{2}\times \dfrac{1}{4})

We need to use the associative property to make it easier.

The associative property for multiplication is as follows :

A\times (B\times C)=(A\times B)\times C

We have,

A=14, B=\dfrac{3}{2}\ and\ C=\dfrac{1}{4}

14(\dfrac{3}{2}\times \dfrac{1}{4})=(14\times \dfrac{3}{2})\times \dfrac{1}{4}

Hence, the correct option is (c).

6 0
3 years ago
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