Question

Find the roots of the quadratic equation:


1/3x2-√11x + 1 =0 ​

Answer 1

Answer:

1/3x^2-√11x+1=0

or, x^2/3-√11x+1=0

( x^2-3√11x+3) /3=0

or, x^2-3√11x+3=0

By applying quadratic formula ,

x=1, y=-3√11x and c=3.

x=(-b_+√b^2-4ac) /2a

=-(-3√11x_+√-3√11x^2-4*1*3) /2*1

=(3√11_+√99-12) /2

=(3√11_+√87) /2

First root=(3√11+√87) /2

=9.638

And, Second root=(3√11-√87) /2

=0.311

I hope it is helpful to you... plzz mark it as brainliest...

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