Math, asked by taha2993, 10 months ago

find the roots of the following quadrick equation using the quardric formula 1/3x2√11x+1=0​

Answers

Answered by digvijaysolanki54
0

Step-by-step explanation:

x=-b+/√bsquare -4 ac ÷2a then by using this formula the answer comes to be -√11+√29/3 ÷2/3 or -√11-√29/3 ÷2/3

Answered by Anonymous
5

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

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