Question

Find roots of the quadratic equation 1/3x^2-11x+1=0​

Answer 1

$\textbf{Given:}$

$\frac{1}{3}x^2-11\,x+1=0$

$\textbf{To find:}$

$\text{The roots of the given equation}$

$\textbf{Solution:}$

$\text{Consider,}$

$\frac{1}{3}x^2-11\,x+1=0$

$\text{This can be written as}$

$x^2-33\,x+3=0$

$\text{We apply quadratic formula to find its roots}$

$\bf\,x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\bf\,x=\dfrac{33\pm\sqrt{(-33)^2-4(1)(3)}}{2(1)}$

$\,x=\dfrac{33\pm\sqrt{1089-12}}{2}$

$\,x=\dfrac{33\pm\sqrt{1077}}{2}$

$\textbf{Answer:}$

$\textbf{The required roots are \bf\dfrac{33\pm\sqrt{1077}}{2} }$

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Answer 2

GIVEN

$\bold{ \bull \frac{1}{3} {x}^{2} - \sqrt{11}x + 1 = 0 }$

FIND

$\bold{ \bull roots \: of \: eq. \: \frac{1}{3} {x}^{2} - \sqrt{11}x + 1 = 0 }$

FORMULA USED

$\bold{ \bull quadratic \: formula = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} }$

SOLUTION

$\bold{ \to\frac{1}{3} {x}^{2} - \sqrt{11}x + 1 = 0 }$

we, know that

$\bold{ \longrightarrow x_{1, 2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} }$

where,

a = $\bold{\frac{1}{3}}$

b = -√11

c = 1

put these values in quadratic formula

$\bold{ ➳ x_{1, 2} = \frac{ -( - \sqrt{11}) ± \sqrt{ {( - \sqrt{11}) }^{2} - 4 (\frac{1}{3})(1) } }{2( \frac{1}{3}) } }$

$\bold{ ➳ x_{1, 2} = \frac{ \not -( \not - \sqrt{11}) ± \sqrt{11 - 4 (0.33333) } }{0.66666 } }$

$\bold{ ➳ x_{1, 2} = \frac{ \sqrt{11} ± \sqrt{11 - 1.33333 } }{0.66666 } }$

$\bold{ ➳ x_{1, 2} = \frac{ \sqrt{11} ± \sqrt{9.66666 } }{0.66666 } }$

$\bold{ ➳ x_{1, 2} = \frac{ \sqrt{11} ± 3.10912 }{0.66666 } }$

$\bold{ \to x_{1} = \frac{ \sqrt{11} + 3.10912 }{0.66666 } }$

$\bold{ \to x_{2} = \frac{ \sqrt{11} - 3.10912 }{0.66666 } }$

$\bold{ \red\to x_{1} = \frac{ \sqrt{11} }{0.66666} + \frac{3.10912}{0.66666} }$

$\bold{ \red\to x_{2} = \frac{ \sqrt{11} }{0.66666} - \frac{3.10912}{0.66666} }$

$\bold{ \blue\to x_{1} = 4.97493 + 4.66368 }$

$\bold{ \blue\to x_{2} = 4.97493 - 4.66368 }$

Hence,

$\bold{x_1}$ = 9.63862

$\bold{x_2}$ = 0.31124