Solve this please .......
1 answer:
Answer:
The sum of reciprocals is 2/3.
You don't need complex numbers to solve this, but if you try to find a and b you will need complex numbers.
Step-by-step explanation:
a+b = 2
a*b = 3
1/a + 1/b = x
(a*b)*(1/a + 1/b) = (a*b)x
b + a = (a*b)(x)
2 = 3x
x = 2/3
b = 2 - a
a*(2 - a) = 3
-a^2 + 2a = 3
-a^2 + 2a - 3 = 0
a^2 - 2a + 3 = 0
let's solve the quadratic equation
a^2 - 2a + 3 = a^2 - 2a + 1 + 2 = (a - 1)^2 + 2 = 0
(a - 1)^2 = -2
these options correspond to a and b from the original question.
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See picture for solution to your problem.
Answer: Any real number x as long as and
In other words, anything but 0 or -2/3 is valid.
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Explanation:
Set the denominator equal to zero and solve for x
2(3x^2 + 2x) = 0
3x^2 + 2x = 0
x(3x + 2) = 0
x = 0 or 3x+2 = 0 .... zero product property
x = 0 or 3x = -2
x = 0 or x = -2/3
If either x = 0 or x = -2/3, then the denominator 2(3x^2 + 2x) will be zero. But recall that we cannot have zero in the denominator. Dividing by zero is not allowed. The expression is undefined when we divide by zero.
Therefore, we must exclude x = 0 and x = -2/3 from the domain. Any other real number is valid as an x input.