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

Steps to Solve

1. Solve for Set A

Starting with the equation $4x^2 = 2x + 3$, we can rearrange it:

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

Next, we will use the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where ( a = 4, b = -2, c = -3 ).

Calculating the discriminant:

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

Now substituting back into the formula:

$$ x = \frac{2 \pm \sqrt{52}}{8} $$

Simplifying further gives:

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

Thus, Set A is:

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

2. Solve for Set B

Starting with the equation $log(x^2) = (log(x))^2$, we can rewrite it using properties of logarithms:

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

Next, letting ( y = \log(x) ), we can rewrite the equation as:

$$ 2y = y^2 $$

Rearranging gives:

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

So,

$$ y = 0 \quad \text{or} \quad y = 2 $$

Now substituting back for ( x ):

1. If ( y = 0 ), then ( \log(x) = 0 ) implies ( x = 1 ).
2. If ( y = 2 ), then ( \log(x) = 2 ) implies ( x = 10^2 = 100 ).

Thus, Set B is:

$$ B = { 1, 100 } $$

3. Find the Intersection A ∩ B

Now we find the common elements between Set A and Set B:

Set A has:

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

Evaluating these values numerically:

• ( \frac{1 + \sqrt{13}}{4} \approx 1.30 )
• ( \frac{1 - \sqrt{13}}{4} \approx -0.30 )

So, Set A approximately contains one element greater than 1 and the other is negative.

Set B has:

$$ B = { 1, 100 } $$

Comparing the elements, we see that:

The only common element is ( 1 ).