Question

3x+ 1/3y = 10/3 find quadratic equations.

Answer 1

$\red{ \huge{ \underline{ \overline{ \mid{ \bold{ \purple{ \: \: SOLUTION\: \: \red{ \mid}}}}}}}}$



$\underline{\underline{\bold{ \: (1) \: \: Calculation \: \: for \: \: finding \: \: the \: \:}}}\\ \underline{\underline{\bold{ \: \: Quadratic \: \:equation \: \: : }}}$




$\bold{3x + \frac{1}{3x} = \frac{10}{3} } \\ \\ \bold{ \Longrightarrow \: \frac{9x {}^{2} + 1}{3x} = \frac{10}{3} } \\ \\ \bold{ \Longrightarrow \: 3(9x {}^{2} + 1) = 10 \times 3x} \\ \\ \bold{ \Longrightarrow \: 27x {}^{2} + 3 = 30x} \\ \\ \bold{ \Longrightarrow \: 27x {}^{2} - 30x + 3 = 0} \\ \\ \bold{ \therefore \: \: \: \underline{ \: \: 9 x {}^{2} - 10 + 1 = 0 \: \: }}$

$\underline{ \bold{ \: \: The \: \: quadratic \: \: equation \: \: is \: \: : }} \\ \\ \: \: \: \: \: \: \: \: \boxed{ \boxed{ \bold{ \red{ \: \: 9x {}^{2} - 10x + 1 = 0 \: \: \: }}}}$



$\underline{\underline{\bold{\:(2)\: \: Calculation \: \: for \: \: finding \: \:the\: \:roots \: \: }}}$



$\bold{ \: \: \: \: \: \: \: \: \: \: \: 9x {}^{2} - 10 + 1 = 0} \\ \\ \bold{ \Longrightarrow \: 9x {}^{2} - 9x - x + 1= 0 } \\ \\ \bold{ \Longrightarrow \: 9x(x - 1) - 1(x - 1) = 0} \\ \\ \bold{ \Longrightarrow \: (x - 1)(9x - 1)= 0}$

$\bold{ \therefore \: \: x - 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \bold{ \Longrightarrow \: \: x = 1}$

$\bold{OR,}$

$\bold{9x - 1 = 0} \\ \\ \bold{ \Longrightarrow \: \: 9x = 1} \\ \\ \bold{ \Longrightarrow \: \: x = \frac{1}{9} }$

$\bold{ \therefore \: \: \: The \: \: roots \: \: of \: \: the \: \:quadratic } \\ \\ \bold{equations \: \: are \: \: : \: \: \red{ \: \: x = 1 \: \: \: \: Or ,\: \: \: \: \: x = \frac{1}{9} }}$

Answer 2

$\underline{ \underline{ \huge{ \: \: SOLUTION \: \: }}}$

$\underline{(1) \: \: Finding \: \: the \: \: quadratic \: \: equation \: : }$

$3x \: + \: \frac{1}{3x} \: = \: \frac{10}{3} \\ \\ = > \frac{(3x \: \times \: 3x) \: + \: 1}{3x} \: = \: \frac{10}{3} \\ \\ = > \frac{9x {}^{2} \: + \: 1 }{3x} = \frac{10}{3} \\ \\ = > 27x {}^{2} + 3 = 30x \\ \\ = > 27x {}^{2} \: - \: 30x \: + 3 \: = \: 0 \\ \\ = > 9x {}^{2} \: - \: 10x\: + \: 1 \: = \: 0$

$Hence, \\ \\ the \: \: required \: \: quadratic \: \: equation \: \: \\ \\ is \: \: : \: \boxed{ \bold{ \: \: 9x {}^{2} - 10x + 1 = 0 \: }}$

$\underline{(2) \: \: Finding \: \: the \: \: roots \: \: of \: \: the \: \: quadratic} \\ \underline{\: \: \: \: \: equation \: \: :}$

$9x {}^{2} - 10x + 1 = 0 \\ \\ = > 9x {}^{2} - 9x - x + 1 = 0 \\ \\ = > 9x(x - 1) - 1(x - 1) = 0 \\ \\ = > (x - 1)(9x - 1) = 0$

$\therefore \: x - 1 = 0 \\ \\ = > x = 1 \\ \\ \\ Or ,\: \\ \\ \\ 9x - 1 = 0 \\ \\ = > x = \frac{1}{9}$