How do you solve #3q ^ { 2} - 38q - 13= 0#?

2 Answers
MysticSmile
Sep 13, 2017

#q = -1/3 or 13#

Explanation:

#3q^2-38q-13q=0#
You have to factor the left side of the equation. #(3q+1)*(q-13)=0# Next, set the factors to 0. #3q+1=0# and #q-13=0#. #q# can either be (fraction) #-1/3# in the first case or it can be #13# in the second case.

EZ as pi
Sep 13, 2017

#q=13 or q=-1/3#

Explanation:

#3q^2 -38q-13=0#

Factorise the quadratic trinomial.

Find factors of #3 and 13# whose products differ by #38#

Note that both #3 and 13# are prime numbers, so there are not many combinations to try.

#" "3 and 13#
#" "darr" "darr#
#" "1" "13" "rarr 3xx13 = 39#
#" "3" "1" "rarr 1 xx 3 rarr = ul1#
#color(white)(xxxxxxxxxxxxxxxxxxx)38larr# the difference

These are the correct factors, now put in the signs:

#" "3 and 13#
#" "darr" "darr#
#" "1" "-13" "rarr 3xx-13 = -39#
#" "3" "+1" "rarr 1 xx +3 rarr = +ul1#
#color(white)(xxxxxxxxxxxxxxxxxxxxxx)-38larr# the difference

#(q-13)(3q+1)=0#

Set each factor equal to #0#

If #q-13=0, rarr q=13#

If #3q+1=0, rarr q = -1/3#

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