How do you solve #a+1/2 + 3/2 = 1/a #?

Question

How do you solve #a+1/2 + 3/2 = 1/a  #?

Savannah Anderson · Accepted Answer

You multiply everything by the same amount. But there is a shortcut in this particular case.

You may have noticed that #1/2and3/2# have the same denominator, so they can be added to #4/2=2# #->a+2=1/a# We now multiply everything by #a#: #->a*a+2*a=1# If we subtract #1# from both sides we get a quadratic equation: #->a^2+2a-1=0# which can be solved.

Abigail Allen · Answer

undefined To solve the equation a + 1/2 + 3/2 = 1/a, we can start by combining the fractions on the left side of the equation. This gives us (a + 1/2 + 3/2) = (2a + 1)/2.

Next, we can simplify the equation by finding a common denominator for the fractions on the left side. The common denominator is 2, so we have (2a + 1)/2 = 1/a.

To eliminate the fractions, we can multiply both sides of the equation by 2a. This gives us (2a + 1) = 2/a.

Expanding the equation, we have 2a + 1 = 2/a.

To get rid of the fraction, we can multiply both sides of the equation by a. This gives us a(2a + 1) = 2.

Expanding the equation again, we have 2a^2 + a = 2.

Rearranging the equation, we have 2a^2 + a - 2 = 0.

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's solve it by factoring.

Factoring the quadratic equation, we have (2a - 1)(a + 2) = 0.

Setting each factor equal to zero, we have 2a - 1 = 0 or a + 2 = 0.

Solving these equations, we find that a = 1/2 or a = -2.

Therefore, the solutions to the equation a + 1/2 + 3/2 = 1/a are a = 1/2 and a = -2.