Solve:
2x^-2 - 5x^-1 - 3 = 0
2a^2 -5a - 3 = 0
(2a^2 - ) + ( a - ) = 0
I am stuck here please help.
is 2a^2 -5a - 3 = 0
supposed to be the simplified version of
2x^-2 - 5x^-1 - 3 = 0 ??
your first equation would be
2/x^2 - 5/x - 3 = 0
multiply each term by x^2
2 - 5x - 3x^2 = 0
3x^2 + 5x - 2 = 0
(3x-1)(x+2) = 0
x = 1/3 or x = -2
yes that is the sinplified version of
To solve the equation 2x^(-2) - 5x^(-1) - 3 = 0, you can start by making a substitution. Let's substitute a = x^(-1).
With this substitution, the equation becomes:
2(a^2) - 5a - 3 = 0
Now, we can try to factor or use the quadratic formula to find the values of a that satisfy this equation.
To factor the quadratic equation, we need to find two numbers whose product is -6 (2*(-3)) and whose sum is -5, the coefficient of 'a'. After trying different pairs of factors, we find that -6 and +1 satisfy these conditions.
Therefore, we can rewrite the equation as:
2(a^2 - 6a + 1a - 3) = 0
Now, group the terms and factor:
2(a(a - 6) + 1(a - 3)) = 0
Next, we use the zero product property, which states that if the product of two numbers is zero, then at least one of the numbers must be zero. So, we set each factor equal to zero and solve for 'a'.
Setting a = 0:
a = 0
Setting (a - 6) = 0:
a - 6 = 0
a = 6
Setting (a - 3) = 0:
a - 3 = 0
a = 3
Therefore, we have three possible solutions for 'a': a = 0, a = 6, and a = 3.
Now, let's substitute back the value of 'a' into the original substitution: a = x^(-1).
For a = 0, we get:
x^(-1) = 0
However, x^(-1) cannot be equal to zero because any number divided by zero is undefined. Therefore, a = 0 is not a valid solution.
For a = 6, we get:
x^(-1) = 6
To solve for 'x', we can take the reciprocal of both sides:
1/x = 6
Now, solve for 'x':
x = 1/6
For a = 3, we get:
x^(-1) = 3
To solve for 'x', we can take the reciprocal of both sides:
1/x = 3
Now, solve for 'x':
x = 1/3
Therefore, the solutions to the equation 2x^(-2) - 5x^(-1) - 3 = 0 are x = 1/6 and x = 1/3.