set everything to zero, then you can factor.
#4
6s^2-17s+12 = 0
You know the factors will look like
(6x-?)(x-?) or (3x-?)(2x-?)
So, since -17 is fairly large, we will probably have the first
(6s-3)(s-4) gives -27s too big
(6s-4)(s-3) gives -22s still too big
(6s-6)(s-2) gives -18s close, but no cigar
(3s-4)(2s-3) gives -17s YES!
Do the others the same way. #1 involves a difference of squares, so it's easy.
I need help on these problems. There over solving equations by factoring.
1. 4x^2=9
2. M^2-36=16M
3. R^2+9=10r
4. 6s^2=17s-12
3 answers
Would #4 be s=1 and s=2?
well, no.
If the product of some numbers is zero, then one of the numbers must be zero. We have
(3s-4)(2s-3) = 0
So, we must have either
3s-4 = 0
or 2s-3 = 0
So, s = 4/3 or 3/2
Had you tried your guesses in the original equation, you would have seen they did not satisfy it.
If the product of some numbers is zero, then one of the numbers must be zero. We have
(3s-4)(2s-3) = 0
So, we must have either
3s-4 = 0
or 2s-3 = 0
So, s = 4/3 or 3/2
Had you tried your guesses in the original equation, you would have seen they did not satisfy it.
2 answers