How do you solve $1/3t^2 + 3= 2t$ using the formula?

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Answer 1 · 2015-06-27T20:24:28

First subtract $2t$ from both sides to get:

$1/3t^2-2t+3 = 0$

$t = (2+-sqrt(2^2-(4xx1/3xx3)))/(2*1/3) = 3$

$1/3t^2-2t+3$ is in the form $at^2+bt+c$ with $a=1/3$ , $b=-2$ and $c=3$

So the roots of $1/3t^2-2t+3 = 0$ are given by the quadratic formula:

$t = (-b+-sqrt(b^2-4ac))/(2a)$

$=(2+-sqrt(2^2-(4xx1/3xx3)))/(2*1/3)$

$=(2+-sqrt(4-4))/(2*1/3)$

$=cancel(2)/(cancel(2)*1/3)$

$=3$

Alternatively, multiply the quadratic equation by $3$ to get:

$t^2-6t+9 = 0$

Notice that the coefficients (ignoring signs) are 1,6,9.

Does the pattern $1, 6, 9$ ring any bells?

Well $169 = 13^2$ and non-coincidentally:

$t^2-6t+9 = (t-3)^2$

$t^2-6t+9 = 0$ when $(t-3) = 0$ , that is when $t=3$

How do you solve 1/3t^2 + 3= 2t1/3t^2 + 3= 2t using the formula?