Question

What is the solution set to the inequality (4x-3)(2x-1)20?

Answer 1

The given question is wrong.

Question:

What is the solution set to the inequality (4x – 3) (2x – 1) ≥ 0?

(A) $\{x| x\leq 3\ \text {or} \ x\geq 1$

(B) $\{x| x\leq 2\ \text {or} \ x\geq \frac{4}{3}$

(C) $\{x| x\leq \frac{1}{2}\ \text {or} \ x\geq \frac{3}{4}$

(D) $\{x| x\leq \frac{-1}{2}\ \text {or} \ x\geq \frac{-3}{4}$

Answer:

The solution set to the given inequality is $\{x| x\leq \frac{1}{2}\ \text {or} \ x\geq \frac{3}{4}$.

Solution:

Given expression is (4x – 3) (2x – 1) ≥ 0.

Let us take the expression is equal to zero.

(4x – 3) (2x – 1) = 0

By quadratic factor, If AB = 0, then A = 0 or B = 0.

(4x – 3) = 0 or (2x – 1) = 0

Let us take the first factor equal to zero.

⇒ 4x – 3 = 0

⇒ 4x = 3

$x=\frac{3}{4}$

Now, take the second factor equal to zero.

⇒ 2x – 1 = 0

⇒ 2x = 1

$x=\frac{1}{2}$

So, $x=\frac{1}{2},x=\frac{3}{4}$.

Now, write it in the inequality to make the statement true.

$x\geq \frac{1}{2}\ \text{(or)}\ x\leq \frac{3}{4}$

Option C is the correct answer.

The solution set to the given inequality is $\{x| x\leq \frac{1}{2}\ \text {or} \ x\geq \frac{3}{4}$.