How do you solve #7a^2+24=-29a#?

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Answer 1 · 2015-06-18T12:09:59

I solve by writing as a quadratic, then factoring. (If factoring hadn't worked quickly, I would have used the quadratic formula.)

#7a^2+24=-29a# if and only if

#7a^2 + 29a+24 = 0#

If it is easily factorable, the factors must look like:

#(7a + color(white)"sss" )(a + color(white)"sss")#

And the spaces are filled by two factors of 24:

#(7a+m)(a+n)#

#m xx n = 24# where #ma+7na = 29a#

#1 xx 24# and #24 xx 1# won't work
#2 xx 12# and #12 xx 2# won't work
#3 xx 8# won't work in that order, but #8 xx 3# works.

Check to be sure
#(7a+8)(a+3) = 7a^2 +21a+8a+24 = 7a^2 +29a +24#

So we have:
#7a^2 + 29a+24 = 0#

#(7a+8)(a+3) = 0#

#7a+8 =0# #" or "# #a+3=0#

#a = -8/7# #" or "# #a = -3#