Question

Consider a equation for t below 191×2=3.97t×t+33.4t
What value of t satisfies this equation

Answer 1

Given that,

$\Rightarrow 191×2=3.97t×t+33.4t$

$\Rightarrow 382=3.97t^2+33.4t$

$\Rightarrow 3.97t^2+33.4t-382=0$

it is the quadratic equation, so it will have 2 roots.

Now, from the quadratic equation formula,

$\Rightarrow t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow t=\dfrac{-33.4\pm\sqrt{(33.4)^2+4\times 3.97\times382}}{2\times 3.97}$

$\Rightarrow t=\dfrac{-33.4\pm\sqrt{1115.56+6066.16}}{7.94}$

$\Rightarrow t=\dfrac{-33.4\pm \sqrt{7181.72}}{7.94}$

$\Rightarrow t=\dfrac{-33.4\pm84.74}{7.94}$

Taking "+",

$\Rightarrow t=\dfrac{-33.4+84.74}{7.94}=6.46$

so,

$t=6.5$

$​​$

Taking "-"

$t=\dfrac{-118.14}{7.94}=-14.88$