Does (6(t^2 -1))/(2(t^2 +1))=3?
[ 6 (t-1) (t+1) ] / [ 2 ( t-i)(t+i) ] is a complex number using i = sqrt(-1)
(6(t^2 -1))/(2(t^2 +1))=3
(6(t^2 -1)) = 3(2(t^2 +1))
6(t^2-1) = 6(t^2+1)
t^2-1 = t^2 + 1
-1 = 1
There is no value of t which will work.
To determine if the equation (6(t^2 - 1))/(2(t^2 + 1)) = 3 is true, we need to simplify the expression on the left side and then compare it to the number 3.
Let's simplify the expression step by step:
1. Start with (6(t^2 - 1))/(2(t^2 + 1)).
2. Distribute the 6 to both terms within the numerator: 6 * t^2 - 6 * 1.
This gives us (6t^2 - 6)/(2(t^2 + 1)).
3. Now, distribute the 2 to both terms within the denominator: 2 * t^2 + 2 * 1.
This gives us (6t^2 - 6)/(2t^2 + 2).
4. Now, let's simplify the numerator by factoring out a common factor of 6:
Divide both terms in the numerator by 6: (6/6)t^2 - (6/6).
This simplifies to t^2 - 1.
5. Similarly, let's simplify the denominator by factoring out a common factor of 2:
Divide both terms in the denominator by 2: (2/2)t^2 + (2/2).
This simplifies to t^2 + 1.
Now, let's substitute the simplified forms back into the original equation:
(t^2 - 1)/(t^2 + 1) = 3.
After simplifying, we no longer have any fractions. So, to determine if the equation is true, we need to check if there is any value of t for which the equation holds.
By multiplying both sides of the equation by (t^2 + 1), we eliminate the denominator:
(t^2 - 1) = 3(t^2 + 1).
Expanding both sides:
t^2 - 1 = 3t^2 + 3.
Further simplifying by combining like terms:
-1 = 2t^2 + 3.
Bringing all terms to one side:
2t^2 + 4 = 0.
Since it is a quadratic equation, we can attempt to solve it by factoring, completing the square, or using the quadratic formula. However, in this particular case, the quadratic equation has no real solutions.
Therefore, the equation (6(t^2 - 1))/(2(t^2 + 1)) = 3 is not true for any real value of t.