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

Question

Isaiah Anthony · Accepted Answer

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

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

Then using the quadratic formula:

#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#

Kennedy Adams · Answer

undefined To solve the equation $ \frac{1}{3}t^2 + 3 = 2t $ using the formula, first rearrange the equation into standard quadratic form, $ at^2 + bt + c = 0 $. Then, apply the quadratic formula, which is given by:

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

where $ a $, $ b $, and $ c $ are the coefficients of the quadratic equation.

For the given equation $ \frac{1}{3}t^2 + 3 = 2t $, rearrange it into standard form:

$$ \frac{1}{3}t^2 - 2t + 3 = 0 $$

Now, identify $ a = \frac{1}{3} $, $ b = -2 $, and $ c = 3 $.

Substitute these values into the quadratic formula:

$$ t = \frac{{-(-2) \pm \sqrt{{(-2)^2 - 4(\frac{1}{3})(3)}}}}{{2(\frac{1}{3})}} $$

Simplify the expression:

$$ t = \frac{{2 \pm \sqrt{{4 - 4}}}}{{\frac{2}{3}}} $$
$$ t = \frac{{2 \pm \sqrt{0}}}{{\frac{2}{3}}} $$
$$ t = \frac{2}{{\frac{2}{3}}} $$

Solve for $ t $:

$$ t = 3 $$

Therefore, the solution to the equation $ \frac{1}{3}t^2 + 3 = 2t $ using the quadratic formula is $ t = 3 $.