How do you solve $\frac{- 2}{2 t + 1} = - t$ $\frac { - 2} { 2t + 1} = - t$ ?

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How do you solve $\frac{- 2}{2 t + 1} = - t$ $\frac { - 2} { 2t + 1} = - t$ ?

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Answer 1 · 2018-05-16T19:53:42

$- \frac{2}{2 t + 1} = (- t)$ $-2/(2t+1)=(-t)$
Multiply the whole equation by $- (2 t + 1)$ $-(2t+1)$ to get rid of the minus sign
$\Rightarrow 2 = t (2 t + 1)$ $rArr2=t(2t+1)$
$\Rightarrow 2 t^{2} + t - 2 = 0$ $rArr2t^2+t-2=0$
Now just fill the quadratic formula
$t_{1, 2} = \frac{- 1 \pm \sqrt{1^{2} - 4 (2 (- 2))}}{2 \cdot 2}$ $t_(1,2)=(-1+-sqrt(1^2-4(2(-2))))/(2*2)$
$t_{1, 2} = \frac{- 1 \pm \sqrt{1 + 16}}{4}$ $t_(1,2)=(-1+-sqrt(1+16))/4$
$t_{1} = \frac{- 1 + \sqrt{17}}{4}$ $t_1=(-1+sqrt17)/4$
$t_{2} = \frac{- 1 - \sqrt{17}}{4}$ $t_2=(-1-sqrt17)/4$

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How do you solve $\frac{- 2}{2 t + 1} = - t$ $\frac { - 2} { 2t + 1} = - t$ ?

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