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scoray [572]
3 years ago
12

What are the solutions of 3x^2-x+4=0

Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0
3x² -x+4=0

We can use quadratic formula
a=3, b= -1, c=4
 x=  \frac{-b+/- \sqrt{b^{2}-4ac} }{2a}  
\\ \\ x= \frac{1+/- \sqrt{1-4*3*4} }{2*3} 
\\ \\ x =  \frac{1+/- \sqrt{-47} }{6} 

This equation does not have real  roots, it has only 2 imaginary roots.

x= \frac{1+ \sqrt{47}i }{6} 
\\ \\ x=  \frac{1- \sqrt{47}i }{6}


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Kruka [31]

Answer:

D. Corey!

Step-by-step explanation:

I hope this helps! Have a great rest of your day!

5 0
3 years ago
What are the solutions of -1 ≤ x – 3 < 4
cricket20 [7]

Answer:

2 ≤ x  < 7

Step-by-step explanation:

-1 ≤ x – 3 < 4

Add 3 to each side

-1+3 ≤ x – 3+3 < 4+3

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7 0
3 years ago
Y = x + 2<br> 2x - y = -4
mrs_skeptik [129]
1.) X=-2

2.) 2x-y+4=0


Hope this helps! :))
3 0
3 years ago
Help will give brainliest!!!!!!
sashaice [31]

Answer:

2.5

type A/B

Step-by-step explanation:

10/2.5=4

15/6=2.5

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7 0
3 years ago
What is the following sum 3b^2
densk [106]

The given expression is 3b^2*(\sqrt[3]{54a}) + 3*(\sqrt[3]{2ab^6})

This can be simplified as :

= 3*b^2*(\sqrt[3]{27 *2*a}) + 3*(\sqrt[3]{2*a*b^6})

We know that: \sqrt[3]{27}  = 3

Similarly we also can simplify: \sqrt[3]{b^6}  = b^2

So our expression will look like this:

= 3*3*b^2*(\sqrt[3]{2a}) + 3*b^2*(\sqrt[3]{2a})

= 9b^2*(\sqrt[3]{2a}) + 3b^2*(\sqrt[3]{2a})

=\sqrt[3]{2a}*(9b^2 + 3b^2)

=\sqrt[3]{2a}*(12b^2)

This can also be written as:

12b^2(\sqrt[3]{2a})

So the Answer is Option B


6 0
3 years ago
Read 2 more answers
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