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Simplified (4x3 - 5x2 + 3x) + (-2x3 - x2 + 6x)

bogdanovich [222]

Answer:

Answer is imaginary roots

Step-by-step explanation:

$({4x}^{3} - {5x}^{2} + 3x) + ( - {2x}^{3} - {x}^{2} + 6x) \\ 2 {x}^{3} - 6 {x}^{2} + 9x = 0 \\ dividing \: by \: x \\ 2 {x}^{2} - 6x + 9 = 0 \\ dividin g\: by \: 2 \\ {x}^{2} - 3x + \frac{9}{2} = 0 \\ shifting \: the \: constant \: term \: on \: right \: side \\ {x}^{2} - 3x = - \frac{9}{2} \\ adding \: \frac{1}{2} \times (coefficient \: of \: x) \: whole \: square \\$

On solving the above mentioned equation we get some imaginary values.

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THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

6 0

3 years ago

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Now subtract b/(2a) on both sides:

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Combine the fractions (they have the same denominator):

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

6 0

3 years ago

GUYS I NEED HELPPPPPP PLEASEEEEEEEEEEE IF YOU CAN THEN TYSM

love history [14]

Answer:

198 cubic metres

Steps:

Find the area of cross section

Area if traingle + Area of square

1/2 × 3 × 4 + 4×4

6+ 16

22

Then times with the lenght (9)

22×9

I hope this helps, dont hesitate to ask for any question.

Mark me as brainliest is appreciated.Tq!!

7 0

3 years ago

PLEASE HELP ASAP!! SPAM ANSWERS WILL BE DELETED

kompoz [17]

Answer:

the answer is B), k = 6j

Step-by-step explanation:

the varible is joanne since... you dont know how many games she won, but you DO know how much games kevin won so:

k = Kevin

j = Joanne (the person you dont know how much times they won)

Hope this helpes!!!! :D

3 0

3 years ago

Help me please :((: thank u

I am Lyosha [343]

Hey there!

In order for you to find the one that is equivalent to the given expression is to COMBINE YOUR LIKE TERMS then work from there!

12q + q^2 - 8 + q = ?

(q^2) + (12q + q) + (-8)

q^2 = q^2 since it doesn't have any like term(s)

12q + q = 12q + 1q = 13q

-8 = -8 since it doesn't have any like term(s)

Put all the numbers we solved for in An equation and Thats your answer!

If you did it correctly you should have: q^2 + 13q - 8

Answer: q^2 + 13q - 8 ✅

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

6 0

3 years ago