I hope your equation t2 + 2t 1 = 0. By inspection put t=0 0^2+2^0 1=0 therfore t=0 is a solution t2 + 2t 1 = 0. 2t = t2 + 1 Taking log on both sides log(2 ...

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Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

Solve t 2+2t-...

Question

Solve: $t^{2} + 2 t + 1 = 0$

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Solution

We have, $t^{2} + 2 t + 1 = 0$
$\Rightarrow t^{2} + t + t + 1 = 0$
$\Rightarrow t (t + 1) + 1 (t + 1) = 0$
$\Rightarrow (t + 1) (t + 1) = 0$
$\Rightarrow (t + 1) = 0 o r, (t + 1) = 0$
$\Rightarrow t = - 1 o r, t = - 1$

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Solve: t2+2t+1=0

We have, t2+2t+1=0⇒t2+t+t+1=0⇒t(t+1)+1(t+1)=0⇒(t+1)(t+1)=0⇒(t+1)=0or,(t+1)=0⇒t=−1or,t=−1

Solve: $t^{2} + 2 t + 1 = 0$

We have, $t^{2} + 2 t + 1 = 0$
$\Rightarrow t^{2} + t + t + 1 = 0$
$\Rightarrow t (t + 1) + 1 (t + 1) = 0$
$\Rightarrow (t + 1) (t + 1) = 0$
$\Rightarrow (t + 1) = 0 o r, (t + 1) = 0$
$\Rightarrow t = - 1 o r, t = - 1$