Math, asked by ATHARGAPALIWAL, 7 months ago

-9.8t^2+9.8t-78.4(solve this by qudratic equation )​

Answers

Answered by pulakmath007
13

\displaystyle\huge\red{\underline{\underline{Solution}}}

 \longmapstoFORMULA TO BE IMPLEMENTED

A general equation of quadratic equation is

a {x}^{2} +  bx + c = 0

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

a {x}^{2} +  bx + c = 0

The roots are given by

 \displaystyle \: x =  \frac{ - b \pm \:  \sqrt{ {b}^{2} - 4ac } }{2a}

 \longmapstoCALCULATION

The given Quadratic Equation is

 - 9.8 {t}^{2}  + 9.8t - 78.4 = 0

Dividing both sides by - 9.8 we get

  {t}^{2}  - t + 6 = 0

Comparing with general equation of quadratic equation

a {t}^{2}  + bt + c = 0

We get

a = 1 \:  ,  \: b \:  =  - 1 \:  , c = 6

So by the Sridhar Acharyya formula the roots are given by

 \displaystyle \: t =  \frac{ - b \pm \:  \sqrt{ {b}^{2} - 4ac } }{2a}

 \displaystyle \: t \:  =  \frac{ 1 \pm \:  \sqrt{ {( - 1)}^{2} - 4 \times 1 \times 6 } }{2 \times 1}

 \implies \:  \displaystyle \: t \:  =  \frac{ 1 \pm \:  \sqrt{  - 23 } }{2}

 \implies \:  \displaystyle \: t \:  =  \frac{ 1 \pm \: i \sqrt{   23 } }{2}

HENCE THE REQUIRED ROOTS ARE

  \displaystyle \:    \frac{ 1  +  \: i \sqrt{   23 } }{2}  \:  \: and \:  \:   \frac{ 1  -  \: i \sqrt{   23 } }{2}

Where i = Imaginary number

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