The number of real solutions of the equation `(9//10)^x=-3+x-x^2` is

Options

A. 2
B. 1
C. 0
D. none of these

Answer

To determine the number of real solutions for the equation

\[ \left(\frac{9}{10}\right)^x = -3 + x - x^2, \]

we will analyze both sides of the equation step by step.

### Step 1: Analyze the Left-Hand Side (LHS)

The left-hand side of the equation is

\[ LHS = \left(\frac{9}{10}\right)^x. \]

Since \(\frac{9}{10}\) is a positive number less than 1, the function \(\left(\frac{9}{10}\right)^x\) is always positive for all real values of \(x\). Therefore, we can conclude:

\[ LHS > 0 \quad \text{for all } x. \]

**Hint:** Remember that any positive base raised to any power will always yield a positive result.

### Step 2: Analyze the Right-Hand Side (RHS)

The right-hand side of the equation is

\[ RHS = -3 + x - x^2. \]

This can be rewritten as

\[ RHS = -x^2 + x - 3. \]

This is a quadratic equation in the standard form \(ax^2 + bx + c\) where \(a = -1\), \(b = 1\), and \(c = -3\). Since \(a < 0\), the parabola opens downwards.

### Step 3: Find the Roots of the Quadratic Equation

To determine the nature of the roots of the quadratic, we calculate the discriminant \(D\):

\[ D = b^2 - 4ac = 1^2 - 4(-1)(-3) = 1 - 12 = -11. \]

Since the discriminant is negative (\(D < 0\)), the quadratic equation has no real roots. This means that the value of the quadratic expression \(-x^2 + x - 3\) does not cross the x-axis and is always less than zero.

### Step 4: Conclusion

Since the left-hand side \(LHS\) is always positive and the right-hand side \(RHS\) is always negative, there are no points where they can be equal. Therefore, the equation

\[ \left(\frac{9}{10}\right)^x = -3 + x - x^2 \]

has no real solutions.

Thus, the number of real solutions is

\[ \boxed{0}. \]