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Vilka [71]
3 years ago
5

What are the solutions to the equation 2x^2 - 2x - 12 = 0?

Mathematics
1 answer:
pantera1 [17]3 years ago
5 0
2(x)^2 - 2(x) - 12 = ?
  
2(x)^2 - 2(x) - 12

(x)^2 - 6 - x

Multiply ↪1 (-6) = -6

1 + -6 = -5 , right?

-3 + 2 = -1, right?

After we have pulled out the like terms we have to add/subtract!   x(x) - 3

Search for the common denominator
2(x) - 3

You can add up 4 for your terms
x + 2 (x - 3)

Make sure you solve this↪ 2 = 0

x + 2 = ?
x + 2 = 0
We have found the first; x = -2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since, this one has a single variable, I'll make the step easier and quicker

Solve this part ↪    x-3 = 0 
If we had the number 3 to your sides we would have found the outcome of the next x
                    
So, the second x is 3 

So, therefore your answer would have to be x =-2 ; x = 3 (most likely option C.)



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6 0
3 years ago
Read 2 more answers
Rationalize the denominator.
Maslowich
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166412

——————————
 
     \mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}


Multiply and divide by the conjugate of the denominator, which is \mathsf{\sqrt{a}+2\sqrt{y}:}

     \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}

The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:
 
     \mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}} 

     \mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

8 0
3 years ago
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