Questions: The function g(x) is defined as g(x)=4^(x^2+2x-15). What 2 real numbers satisfy g(x)=1? A. 0 and 3 B. -3 and 3 C. -3 and 0 D. -5 and 3 E. -5 and 0

Transcript text: The function $g(x)$ is defined as $g(x)=4^{x^{2}+2 x-15}$. What 2 real numbers satisfy $g(x)=1$ ? A. 0 and 3 B. -3 and 3 C. -3 and 0 D. -5 and 3 E. -5 and 0

Solution

Solution Steps

To solve for the real numbers that satisfy $ g(x) = 1 $, we need to set the expression inside the exponent of the base 4 to zero because $ 4^0 = 1 $. Therefore, we solve the equation $ x^2 + 2x - 15 = 0 $. This is a quadratic equation, and we can find the solutions using the quadratic formula.

Step 1: Set the Equation

We start with the function defined as $ g(x) = 4^{x^2 + 2x - 15} $. To find the values of $ x $ that satisfy $ g(x) = 1 $, we set the exponent equal to zero: \[ x^2 + 2x - 15 = 0 \]

Step 2: Solve the Quadratic Equation

Next, we solve the quadratic equation $ x^2 + 2x - 15 = 0 $ using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $ a = 1 $, $ b = 2 $, and $ c = -15 $.

Step 3: Calculate the Roots

Calculating the discriminant: \[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 \] Now substituting back into the quadratic formula: \[ x = \frac{-2 \pm \sqrt{64}}{2 \cdot 1} = \frac{-2 \pm 8}{2} \] This gives us two solutions: \[ x_1 = \frac{6}{2} = 3 \quad \text{and} \quad x_2 = \frac{-10}{2} = -5 \]