Social Sciences, asked by mohammedabdulkhaliq1, 2 months ago

16x *52xsquare*125xcube

going to mark brainliest for the correct answer​

Answers

Answered by YuriTapsuki
0

Answer:

Step 1: Isolate the radical term.

Before we take the square of both sides, we need to isolate the radical term.

\begin{aligned}6+3x&=\sqrt{2x+12}+2x\\\\ 6+x&=\sqrt{2x+12}\\\\ \end{aligned}  

6+3x

6+x

​  

 

=  

2x+12

​  

+2x

=  

2x+12

​  

 

​  

 

Step 2: Take the square of both sides.

Now that we have isolated the radical term, we can square both sides to eliminate the radical:

\begin{aligned}6+x&=\sqrt{2x+12}\\\\ \left(6+x\right)^{\blueD{2}}&=\left(\sqrt{2x+12}\right)^{\blueD{2}}\\\\ 36+12x+x^2&=2x+12\\\\ x^2+10x+24&=0\\\\ (x+6)(x+4)&=0 \end{aligned}  

6+x

(6+x)  

2

 

36+12x+x  

2

 

x  

2

+10x+24

(x+6)(x+4)

​  

 

=  

2x+12

​  

 

=(  

2x+12

​  

)  

2

 

=2x+12

=0

=0

​  

 

We found two possible solutions: x=-6x=−6x, equals, minus, 6 and x=-4x=−4x, equals, minus, 4.

Step 3: Check for extraneous solutions.

The original equation is 6+3x=\sqrt{2x+12}+2x6+3x=  

2x+12

​  

+2x6, plus, 3, x, equals, square root of, 2, x, plus, 12, end square root, plus, 2, x.

When \maroonD{x}=\maroonD{-6}x=−6start color #ca337c, x, end color #ca337c, equals, start color #ca337c, minus, 6, end color #ca337c, we get:

\begin{aligned}6+3\maroonD{x}&=\sqrt{2\maroonD{x}+12}+2\maroonD{x}\\\\ 6+3(\maroonD{-6})&\stackrel{?}=\sqrt{2(\maroonD{-6})+12}+2(\maroonD{-6})\\\\ -12&\stackrel{?}=\sqrt{0}-12\\\\ -12&\stackrel{?}=0-12\\\\ -12&=-12 \end{aligned}  

6+3x

6+3(−6)

−12

−12

−12

​  

 

=  

2x+12

​  

+2x

=

?

 

2(−6)+12

​  

+2(−6)

=

?

 

0

​  

−12

=

?

0−12

=−12

​  

 

So x=-6x=−6x, equals, minus, 6 is indeed a solution.

When \maroonD{x}=\maroonD{-4}x=−4start color #ca337c, x, end color #ca337c, equals, start color #ca337c, minus, 4, end color #ca337c, we get:

\begin{aligned}6+3\maroonD{x}&=\sqrt{2\maroonD{x}+12}+2\maroonD{x}\\\\ 6+3(\maroonD{-4})&\stackrel{?}=\sqrt{2(\maroonD{-4})+12}+2(\maroonD{-4})\\\\ -6&\stackrel{?}=\sqrt{4}-8\\\\ -6&\stackrel{?}=2-8\\\\ -6&=-6 \end{aligned}  

6+3x

6+3(−4)

−6

−6

−6

​  

 

=  

2x+12

​  

+2x

=

?

 

2(−4)+12

​  

+2(−4)

=

?

 

4

​  

−8

=

?

2−8

=−6

​  

 

So x=-4x=−4x, equals, minus, 4 is a solution too!

Conclusion

We found that this equation has two solutions! Both x=-6x=−6x, equals, minus, 6 and x=-4x=−4x, equals, minus, 4 solve the equation.

Explanation:

Hope it helps you :)

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