Question

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

Answer 1

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

$\longmapsto$FORMULA 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

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

$\longmapsto$CALCULATION

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