If A = {x ∈ ℝ : 4x² = 2x + 3} B = {x ∈ ℝ : log x² = (log x)²} find A ∩ B

Steps to Solve

1. Solve the equation for set A

We begin with the equation for set A:

$$ 4x^2 = 2x + 3 $$

Rearranging it, we get:

$$ 4x^2 - 2x - 3 = 0 $$

Using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) where ( a = 4, b = -2, c = -3 ):

$$ b^2 - 4ac = (-2)^2 - 4(4)(-3) = 4 + 48 = 52 $$

Now we can find the roots:

$$ x = \frac{2 \pm \sqrt{52}}{2 \cdot 4} = \frac{2 \pm 2\sqrt{13}}{8} = \frac{1 \pm \sqrt{13}}{4} $$

So, the solutions for set A are:

$$ A = \left{ \frac{1 + \sqrt{13}}{4}, \frac{1 - \sqrt{13}}{4} \right} $$

2. Solve the logarithmic equation for set B

For set B, we have:

$$ \log x^2 = (\log x)^2 $$

Rewriting it gives:

$$ 2\log x = (\log x)^2 $$

Let's set ( y = \log x ). Then we have:

$$ 2y = y^2 $$

Rearranging leads to:

$$ y^2 - 2y = 0 $$

Factoring gives:

$$ y(y - 2) = 0 $$

Thus, ( y = 0 ) or ( y = 2 ).

Returning to the variable ( x ):

1. For ( y = 0 ):

$$ \log x = 0 \Rightarrow x = 10^0 = 1 $$

2. For ( y = 2 ):

$$ \log x = 2 \Rightarrow x = 10^2 = 100 $$

So, the solutions for set B are:

$$ B = { 1, 100 } $$

3. Find the intersection ( A \cap B )

We now need to find common elements in sets A and B.

Set A has:

$$ A = \left{ \frac{1 + \sqrt{13}}{4}, \frac{1 - \sqrt{13}}{4} \right} $$

Set B has:

$$ B = { 1, 100 } $$

Calculating ( \frac{1 + \sqrt{13}}{4} ) approximately:

$$ \sqrt{13} \approx 3.605 $$ $$ \frac{1 + 3.605}{4} \approx \frac{4.605}{4} \approx 1.15125 $$

Calculating ( \frac{1 - \sqrt{13}}{4} ) approximately:

$$ \frac{1 - 3.605}{4} \approx \frac{-2.605}{4} \approx -0.65125 $$

Since ( A ) contains elements approximately ( 1.15125 ) and ( -0.65125 ) while ( B ) contains ( 1 ) and ( 100 ), the only common member is:

$$ 1 \in B $$

Thus, the intersection ( A \cap B = { 1 } $$