Answer:
- 2 negative real solutions
- b = -1/2, -3/2
Step-by-step explanation:
The equation can be put into standard form by subtracting 4:
4b^2 +8b +3 = 0
Number of Solutions
There are no changes in sign of the coefficients of this equation. They are (+ + +), so there are no positive real roots. If we change the sign of the odd-degree term (8b), then the signs are (+ - +), and there are two changes of sign. This indicates there are 0 or 2 negative real solutions.
The discriminant is b²-4ac = 8² -4(4)(3) = 64 -48 = 16. This is positive, indicating two real solutions. The two real solutions are both negative.
Factoring
Now, it can be factored by looking for two factors of 4×3 = 12 that have a sum of 8. Those would be 2 and 6. Then the factorization is ...
(4b +2)(4b +6)/4 = 0 . . . . using 2 and 6 in the binomial terms*
(2b +1)(2b +3) = 0 . . . . . . removing the excess factor of 4
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The solution is the set of values of b that make these factors be zero.
2b +1 = 0
b = -1/2
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2b +3 = 0
b = -3/2
The solutions are b = -1/2 and b = -3/2.
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* More detail on the factorization method:
If you look at the product ...
(ax +b)(cx +d) = acx^2 +(ad +bc)x +bd
you can see that the product of the x^2 coefficient and the constant will be ...
(ac)(bd) = abcd.
This can be factored as (ad)(bc), so that the coefficient of x is the product of two factors of the first and last coefficient. However, to get from this knowledge to the actual polynomial factors requires an additional step.
One way to do this is to write the polynomial factors as ...
(ac·x +ad)(ac·x +bc) = 0
When we multiply this out, we find there is an extra factor of ac that needs to be removed. So, we write the factorization as ...
(acx +ad)(acx +bc)/ac = 0
(a(cx +d))(c(ax +b))/ac = 0
This way, we can see that one of the factors of ac can be removed from the first binomial, and the other factor of ac can be removed from the second binomial. That's what we did above.
(ac)(cx +d)(ax +b)/ac = 0
(cx +d)(ax +b) = 0 . . . . . . . our factorization
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We used ac=4, bd=3, ad=2, bc=6 ⇒ a=2, b=3, c=2, d=1.
