Answer:
C) Both
Step-by-step explanation:
The given equation is:
![0=(3x+2)(x-4)](https://tex.z-dn.net/?f=0%3D%283x%2B2%29%28x-4%29)
To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.
Using this property:
![(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}](https://tex.z-dn.net/?f=%283x%2B2%29%20%3D%200%20%5CRightarrow%20%5Cbold%7Bx%20%3D%20-%5Cfrac%7B2%7D%7B3%7D%7D%5C%5C%28x-4%29%20%3D%200%20%5CRightarrow%20%5Cbold%7Bx%20%3D%204%7D)
So, Erik's solution strategy would work.
Now, let us discuss about Caleb's solution strategy:
Multiply
i.e.
= ![3x^2-10x-8](https://tex.z-dn.net/?f=3x%5E2-10x-8)
So, the equation becomes:
![0=3x^2-10x-8](https://tex.z-dn.net/?f=0%3D3x%5E2-10x-8)
Comparing this equation to standard quadratic equation:
![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
a = 3, b = -10, c = -8
So, this can be solved using the quadratic formula.
![x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-%28-10%29%5Cpm%5Csqrt%7B%28-10%29%5E2-4%5Ctimes3%20%5Ctimes%20%28-8%29%7D%7D%7B2%5Ctimes%203%7D%5C%5Cx%3D%5Cdfrac%7B-%28-10%29%5Cpm%5Csqrt%7B196%7D%7D%7B6%7D%5C%5Cx%3D%5Cdfrac%7B10%5Cpm14%7D%7B6%7D%20%5C%5C%5CRightarrow%20x%3D%204%2C%20-%5Cdfrac%7B2%7D%7B3%7D)
The answer is same from both the approaches.
So, the correct answer is:
C) Both