Question

Which of the following are solutions to the quadratic equation below? Check all that apply.

Answer 1

Given the equation:

`2x^2-4x-3=x\text{ ----- equation 1}`

To solve for x,

step 1: Subtract x from both sides of the equation.

This gives

`\begin{gathered} 2x^2-4x-x-3=x-x \\ \Rightarrow2x^2-5x-3=0\text{ ----- equation 2} \end{gathered}`

Step 2: From equation 2, solve for x by factorization.

Thus, we have

`\begin{gathered} 2x^2-6x+x-3=0 \\ Group\text{ }the\text{ terms,} \\ (2x^2-6x)+(x-3)=0 \\ factor\text{ out the common terms,} \\ 2x(x-3)-1(x-3)=0 \\ \Rightarrow(2x-1)(x-3)=0 \\ thus, \\ 2x-1=0\text{ or x-3 =0} \\ when \\ 2x-1=0 \\ x=-(1)/(2) \\ when \\ x-3=0 \\ x=3 \end{gathered}`

Thus, the solution to the quadratic equation is

`-(1)/(2)\text{ or 3}`

The correct option is