Question

if the quadratic equation 3x+1/x=x+3 written in standard form then
1)a=3,b=3,c=1
2)a=3,b=-3,c=1
3)a=2,b=-3,c=1
4)a=3,b=-3,c=3​

Answer 1

$\textbf{Given:}$

$\mathrm{3x+\dfrac{1}{x}=x+3}$

$\textbf{To find:}$

$\text{Standard form the given equation}$

$\textbf{Solution:}$

$\text{Consider,}$

$\mathrm{\dfrac{3x^2+1}{x}=x+3}$

$\text{Cross multiply, we get}$

$\mathrm{3x^2+1=x(x+3)}$

$\mathrm{3x^2+1=x^2+3x}$

$\text{Rearranging terms, we get}$

$\mathrm{2x^2-3x+1=0}$

$\text{Comparing with}\;\mathrm{ax^2+bx+c=0}$

$\text{we get}$

$\textbf{a=2, b=-3, c=1}$

$\textbf{Answer:}$

$\textbf{option(c) is correct}$

Answer 2

Answer:

Right Answer is option 3

Step-by-step explanation:

\textbf{Given:}Given:

\mathrm{3x+\dfrac{1}{x}=x+3}3x+

x

1

=x+3

\textbf{To find:}To find:

\text{Standard form the given equation}Standard form the given equation

\textbf{Solution:}Solution:

\text{Consider,}Consider,

\mathrm{\dfrac{3x^2+1}{x}=x+3}

x

3x

2

+1

=x+3

\text{Cross multiply, we get}Cross multiply, we get

\mathrm{3x^2+1=x(x+3)}3x

2

+1=x(x+3)

\mathrm{3x^2+1=x^2+3x}3x

2

+1=x

2

+3x

\text{Rearranging terms, we get}Rearranging terms, we get

\mathrm{2x^2-3x+1=0}2x

2

−3x+1=0

\text{Comparing with}\;\mathrm{ax^2+bx+c=0}Comparing withax

2

+bx+c=0

\text{we get}we get

\textbf{a=2, b=-3, c=1}a=2, b=-3, c=1

\textbf{Answer:}Answer:

\textbf{option(c) is correct}option(c) is correct

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