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Anna [14]
4 years ago
6

Find the solution of the square root of the quantity of x plus 3 plus 4 equals 6, and determine if it is an extraneous solution

Mathematics
2 answers:
Darina [25.2K]4 years ago
7 0

Answer:

x=1

Step-by-step explanation:

We have been given an equation \sqrt{x+3}+4=6. We are asked to find the solution of our given equation.

\sqrt{x+3}+4-4=6-4

\sqrt{x+3}=2

Now, we will square both sides of our given equation.

x+3=2^2

x+3=4

x+3-3=4-3

x=1

To see whether x=1 is an extraneous solution or not, we will substitute x=1 in our given equation as:

\sqrt{1+3}+4=6

\sqrt{4}+4=6

2+4=6

6=6

Since both sides of our given equation are equal, therefore, x=1 is a solution for our given equation.

Masteriza [31]4 years ago
4 0
X=1 non extraneous i hope this helps
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Answer:

j = -3, -4

Step-by-step explanation:

j^2 + 7j + 12 = 0

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Step-by-step explanation:

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If a and b are odd integers, then a – b is an even integer.
Readme [11.4K]
So the answer choices are
1. a-b=even
2. a and b are not odd
3. a and b are odd
4. a-b=even
5. a-b=not even

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basically choice 4 and 1 are the same so we cross one out
so
the problem said that a and b are odd so therefor choice2 is wrong and choice 3 is correct

then both are odd
odd-odd=even because
an even number is represented as 2n where n is an integer
an odd number can be represented as 2n+1 so assume you have 2 odd numbers 2 away from each other so odd and (odd+2)
odd+2-odd=2n+1+2-(2n+1)=2n+1+2-2n-1=2n-2n+1-1+2=2
you are left with odd
using integers
7 and 11
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even

so odd-odd=even, it depends on weather you consider 0 odd or even


so the asnwers are:
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Can someone please help me on number 16-ABC
melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

-2x\:

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)>10\left(-1\right)

Simplify

2x>-10

Divide both sides by 2

\frac{2x}{2}>\frac{-10}{2}

x>-5

-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

-2x>-6

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)

Simplify

2x

Divide both sides by 2

\frac{2x}{2}

x

-6

Thus, the two intervals:

\left(-\infty \:,\:3\right)

\left(-5,\:\infty \:\right)

The intersection of these two intervals would be the solution to both inequalities.

\left(-\infty \:,\:3\right)  and \left(-5,\:\infty \:\right)

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR  -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0
3 years ago
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