Question 235477: Solve for a and b
0=a^2 - b^2
1=2ab
Answer by MathPro(14)   (Show Source):
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Your problem is:
Solve for a and b
0=a^2 - b^2 equation (1)
1=2ab equation (2)
Divide both sides of equation (2) by 2a to isolate b. Note you could also divide it by b instead, it doesn’t matter which variable you choose.
1/(2a) = 2ab/(2a)
1/(2a) = b equation (3)
Now take equation (3) and substitute for b in equation (1) to get:
a^2 - [1/(2a)]^2 = 0 equation (4)
get rid of the fraction by multiplying both sides of equation (4) by (2a)^2 .
a^2 * (2a)^ 2 – 1 = 0(2a)^2 = 0
a^2 * 4a^2 – 1 = 0
4a^2*a^2 – 1 =0
use the product rule: x^a * x^b = x ^ (a+b) to get:
4a^4 – 1 = 0
Add 1 to each side
4a^4 = 1
Divide each side by 4:
a^4 = ¼
use this general rule (x^a)^b = x ^a*b:
raise both sides to the power ¼:
(a^4)^(1/4) = (1/4)^1/4
a^(4*1/4) = (1/4)^ ¼
a = (1/4)^1/4
use the general rule :1 ^ x = 1 to get:
a = (1^1/4) /(4^1/4) = 1/ (4^1/4)
use the general rule: 1/x^a = x^-a to get:
a = 4 ^ -¼ equation (5)
lets rewrite this differently by using 4 = 2*2 = 2^2 to get a =( 2^2)^-1/4 which can be rewritten as
a = 2^2*(-1/4)
a = 2 ^-1/2 equation (5)
from equation (1) we have a^2 = b^2 so b must b the same as a
but let’s double check this by getting b from equation (2)
substitute for a from equation (5) in equation (2) to get b:
1=2*2^-1/2*b
1 = 2 ^ 1 * 2 ^ -1/2 *b
1 = 2^ (1 -1/2) *b by the product rule
1 = 2 ^ ½ *b
Divide both sides by 2^1/2 to get:
b = 1/(2^1/2) = 2^-1/2 equation (6)
Our answer is a = b = 2 ^ -1/2
**Note: we could have also gotten this by inspection. If a=b from equation (1) and ½ =a*b from equation (2) then [a=sqrt(1/2) times b=sqrt(1/2)] could be the only solution that would produce the value ½ in ½ = a*b
Good Luck!
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