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tensa zangetsu [6.8K]
3 years ago
7

Which of the following expressions is equal to 3x^2+27

Mathematics
1 answer:
nlexa [21]3 years ago
5 0

<u>Answer:</u>

The roots of the the given equation are +3i or -3i.

<u>Solution:</u>

Given, 3x^2+27=0

First of all compare the given equation with the standard form, i.e ax^2+bx+c=0

On comparing,  

a=3

b=0

c=27

According to the quadratic formula,

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

b^2-4ac= 0-4\times3\times27=-324

As b^2-4ac comes out to be negative, the given equation does not have any real roots,

\sqrt {b^2-4ac}= \sqrt{(-324)}

\sqrt {b^2-4ac}= \sqrt{(324\times -1)}

\Rightarrow \sqrt{324}\times \sqrt{-1}

Root of 324 is 18,

\Rightarrow 18\times i (Because root of -1 is i)

x=\frac{(-0+-18i)}{2\times3}

x=\frac{\pm18i}{6}

x=\frac{+18}{6} \ or \frac{-18}{6}

On dividing 18 and 6 we get,

\therefore x=+3i \ or -3i

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Vadim26 [7]
I haven't learned antiderivitives yet but I can try to logic it

<span>First find f′ and then find f. f′′(x)=3x^3+6x^2−x+2, f′(1)=9, f(1)=−7.

we reverse chain rule

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</span>
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now reverse drivitive again

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</span>
<span>f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x
try evaluating it for x=1
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</span>
<span><span>f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x-13 and 7/30

</span>


</span>ANSWER
<span>f'(x)=3/4x^4+2x^3-1/2x^2+2x+19/4
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