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algol13
3 years ago
12

4x2 + x = -3 how do people get an "i" term ?

Mathematics
1 answer:
Gnesinka [82]3 years ago
3 0
4x²<span> + x = -3

You need to use the quadratic formula to solve this. First make sure to get everything on one side so the equation equals 0.
</span>4x² + x + 3 = 0

x = (-b +/-√b² - 4ac)/2a
x = (-1 +/-√1² - 4(4)(3))/2(4)
x = (-1 +/-√1 - 48)/8
x = (-1 +/-√-47)/8

Now before you continue, remember that you can't have a negative number in the square root. √-47 is the same as √47 × √-1. 
√-1 is the same thing as i, so you can take that out.
It will now look like this: i√47
So now you can continue doing the problem.

x = (-1 +/- i√47)/8
x = -1/8 +/- (i√47)/8
This can't really be simplified further.

So if you are meant to put it in a + bi form, then the answers would be:
x = -1/8 + (i√47)/8
x = -1/8 - (i√47)/8
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Find dy/dx if y =x^3+5x+2/x²-1
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<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

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  • g(x) = x^2-1

Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

7 0
2 years ago
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