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sleet_krkn [62]
3 years ago
11

8x^3 -10x^2 +3x=0 please solve for x

Mathematics
2 answers:
gregori [183]3 years ago
8 0
\bf 8x^3-10x^2+3x=0\implies x(8x^2-10x+3)=0
\\\\\\
x(2x-1)(4x-3)=0\implies 
\begin{cases}
x=0\implies &x=0\\\\
2x-1=0\implies 2x=1\implies &x=\frac{1}{2}\\\\
4x-3=0\implies 4x=3\implies &x=\frac{3}{4}
\end{cases}
serg [7]3 years ago
4 0
So x would equal to x=0, 3/4, 1/2
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Step-by-step explanation:

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3 years ago
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Finger [1]

Answer:

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Step-by-step explanation:

Consider linear differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)

It's solution is of form y\,I.F=\int I.F\,q(x)\,dx where I.F is integrating factor given by I.F=e^{\int p(x)\,dx}.

Given: \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x

We can write this equation as \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x

On comparing this equation with \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x), we get p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x

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we get solution as follows:

\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C

{ formula used: \int \cos x\,dx=\sin x }

Applying condition:y(\pi)=\pi^2

\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C

So, we get solution as :

\frac{y}{x^2}=\sin x+\pi

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3 years ago
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A literal equation is an equation which comprises basically of letters., where letters are nothing but variables.

All the kinds of formulas are said to be examples of literal equations.

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