How do you solve #a^2+a-30=0#?

2 Answers
Jul 8, 2015

Answer:

Solve #y = x^2 + x - 30 = 0#

Explanation:

#y = x^2 + x - 30 = 0#
Find 2 numbers knowing sum (-1) and product (-30). Roots have opposite signs.
Factor pairs of (-30) --> ....(-3, 10)(-5, 6). This sum is 1 = b.
Then the 2 real roots are the opposite: #5 and -6#

Jul 8, 2015

Answer:

Factor the trinomial into two binomials. Set each binomial equal to zero and solve for #a#.

Explanation:

#a^2+a-30=0#

Find two numbers that when added equal #1#, and when multiplied equal #-30#.

The numbers #-5# and #6# fit the pattern.

#a^2+a-30=0# =

#(a-5)(a+6)=0#

Set #(a-5)# equal to zero and solve for #a#.

#a-5=0#

#a=5#

Set #(a+6)# equal to zero and solve for #a#.

#a+6=0#

#a=-6#

#a=-6, 5#