18. Which value(s) of \( x \) satisfies the equation below? \( a: 5 b: 12 c:-10 \) \[ \left(5 x^{2}+2 x\right)\left(2 x^{2}+12 x-10=0\right. \] a. No real zeros b. \( \{-0.7,3.1\} \) c. \( \{0.7,-3.1\} \) d. \( \{33.2,-35.6\} \)

To solve the equation \[ (5x^2 + 2x)(2x^2 + 12x - 10) = 0, \] we need to find the values of \( x \) that satisfy either \( 5x^2 + 2x = 0 \) or \( 2x^2 + 12x - 10 = 0 \). ### Step 1: Solve \( 5x^2 + 2x = 0 \) We can factor out \( x \): \[ x(5x + 2) = 0. \] This gives us two solutions: 1. \( x = 0 \) 2. \( 5x + 2 = 0 \) which simplifies to \( x = -\frac{2}{5} = -0.4 \). ### Step 2: Solve \( 2x^2 + 12x - 10 = 0 \) We will use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 2 \), \( b = 12 \), and \( c = -10 \). Calculating the discriminant: \[ b^2 - 4ac = 12^2 - 4 \cdot 2 \cdot (-10) = 144 + 80 = 224. \] Now, substituting into the quadratic formula: \[ x = \frac{-12 \pm \sqrt{224}}{2 \cdot 2} = \frac{-12 \pm 2\sqrt{56}}{4} = \frac{-12 \pm 2\sqrt{4 \cdot 14}}{4} = \frac{-12 \pm 4\sqrt{14}}{4} = -3 \pm \sqrt{14}. \] ### Step 3: Calculate the approximate values Calculating \( \sqrt{14} \): \[ \sqrt{14} \approx 3.74. \] Thus, the two solutions are: 1. \( x \approx -3 + 3.74 \approx 0.74 \) 2. \( x \approx -3 - 3.74 \approx -6.74 \) ### Summary of Solutions From both parts, we have the solutions: 1. From \( 5x^2 + 2x = 0 \): \( x = 0 \) and \( x = -0.4 \). 2. From \( 2x^2 + 12x - 10 = 0 \): \( x \approx 0.74 \) and \( x \approx -6.74 \). ### Final Values The approximate values of \( x \) that satisfy the equation are: - \( x \approx 0 \) - \( x \approx -0.4 \) - \( x \approx 0.74 \) - \( x \approx -6.74 \) Now, let's check the options provided: - a. No real zeros - b. \( \{-0.7, 3.1\} \) - c. \( \{0.7, -3.1\} \) - d. \( \{33.2, -35.6\} \) None of the options match the calculated values. Therefore, the correct answer is: **a. No real zeros**.