Question

For $x \in R$, the number of real roots of the equation $3 x^{2}-4\left|x^{2}-1\right|+x-1=0$ is ______

Answer 1

Correct Answer: 4

Step 1: Given Data
We are given the following data:
- The equation is \( 3x^2 – 4|x^2 – 1| + x – 1 = 0 \)
- The domain for \( x \) is \( x \in [-1, 1] \).

Step 2: Substituting for \( |x^2 - 1| \)
Using \( |x^2 - 1| = -(x^2 - 1) \), the equation becomes:
\( 3x^2 – 4(-x^2 + 1) + x – 1 = 0 \)
Simplifying the equation:
\( 3x^2 + 4x^2 – 4 + x – 1 = 0 \)

Step 3: Simplifying the Equation
Combine like terms:
\( 7x^2 + x – 5 = 0 \)
Now, we have the quadratic equation:
\( 7x^2 + x – 5 = 0 \)

Step 4: Solving the Quadratic Equation
Using the quadratic formula, the solution is given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, \( a = 7 \), \( b = 1 \), and \( c = -5 \). Substituting these values into the quadratic formula:
\( x = \frac{-1 \pm \sqrt{1^2 - 4(7)(-5)}}{2(7)} \)
Simplifying the discriminant:
\( x = \frac{-1 \pm \sqrt{1 + 140}}{14} \)
\( x = \frac{-1 \pm \sqrt{141}}{14} \)

Step 5: Case for \( x \in (-∞, -1) ∪ (1, ∞) \)
For \( x \in (-∞, -1) ∪ (1, ∞) \), the equation becomes:
\( x^2 - 4(x^2 - 1) + x - 1 = 0 \)
Simplifying the equation:
\( x^2 - x - 3 = 0 \)

Step 6: Solving the New Quadratic Equation
Again, using the quadratic formula with \( a = 1 \), \( b = -1 \), and \( c = -3 \), we solve for \( x \):
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} \)
Simplifying the discriminant:
\( x = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2} \)

Step 7: Total Number of Solutions
From both cases, we have 2 solutions from the first case and 2 solutions from the second case, giving us a total of 4 solutions.
Hence, the total number of solutions is \( \boxed{4} \).