Question

The number of real solutions of the equation \[ \left( \frac{9}{10} \right) = -3 + x - x^2 \] is:

• A. 0
• B. 1
• C. 2
• D. None of these

Hint:

If the discriminant of a quadratic equation is negative, the equation has no real solutions.

Answer 1

The Correct Option is A

Step 1: Rearranging the equation.
The given equation is: \[ \left( \frac{9}{10} \right) = -3 + x - x^2. \] First, bring all terms to one side of the equation: \[ x - x^2 - 3 - \frac{9}{10} = 0. \] Simplify the constants: \[ x - x^2 - \frac{39}{10} = 0. \] Multiply the entire equation by 10 to eliminate the fraction: \[ 10x - 10x^2 - 39 = 0. \] Rearrange the terms: \[ -10x^2 + 10x - 39 = 0. \] Multiply the equation by -1 to simplify the signs: \[ 10x^2 - 10x + 39 = 0. \]
Step 2: Discriminant analysis.
For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is given by: \[ \Delta = b^2 - 4ac. \] For our equation \( 10x^2 - 10x + 39 = 0 \), we have: \[ a = 10, \, b = -10, \, c = 39. \] Now, calculate the discriminant: \[ \Delta = (-10)^2 - 4(10)(39) = 100 - 1560 = -1460. \]
Step 3: Conclusion.
Since the discriminant \( \Delta \) is negative, the quadratic equation has no real solutions. Therefore, the number of real solutions is 0. The correct answer is (a) 0.