How do you solve 3q ^ { 2} - 38q - 13= 03q238q13=0?

2 Answers
Sep 13, 2017

q = -1/3 or 13q=13or13

Explanation:

3q^2-38q-13q=03q238q13q=0
You have to factor the left side of the equation. (3q+1)*(q-13)=0(3q+1)(q13)=0 Next, set the factors to 0. 3q+1=03q+1=0 and q-13=0q13=0. qq can either be (fraction) -1/313 in the first case or it can be 1313 in the second case.

Sep 13, 2017

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

Explanation:

3q^2 -38q-13=03q238q13=0

Factorise the quadratic trinomial.

Find factors of 3 and 133and13 whose products differ by 3838

Note that both 3 and 133and13 are prime numbers, so there are not many combinations to try.

" "3 and 13 3and13
" "darr" "darr
" "1" "13" "rarr 3xx13 = 39 1 13 3×13=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