Question

Solve each of the following
a) \(x(2 x-6)=0\)
b) \(m^2+3 m=4\)
c) \(x^2-49=0\)
d) \(x+2-x-3=0\)

Answer 1

a) $x(2x-6)=0$ can be factored as $2x(x-3)=0$. Therefore, the solutions are $x=0$ and $x=3$.

 

b) Rearranging the equation gives $m^2+3m-4=0$. Factoring this gives $(m+4)(m-1)=0$. Therefore, the solutions are $m=-4$ and $m=1$.

 

c) Rearranging the equation gives $(x+7)(x-7)=0$. Therefore, the solutions are $x=-7$ and $x=7$.d) Simplifying the left side gives $-1=0$, which is not true. Therefore, there are no solutions to this equation.