Respuesta :
We have to check which one of the numbers given are solutions of the equation:
[tex]\begin{gathered} 2x^2+9x-3=-x^2+x \\ 2x^2+x^2+9x-x-3=0 \\ 3x^2+8x-3=0 \end{gathered}[/tex]For doing so, we will replace each one of the values onto the equation, and if both sides of the equation remains the same, we will get a solution.
a: x=-2
In this case,
[tex]\begin{gathered} 3(-2)^2+8(-2)-3=3(4)+16-3 \\ =12-16-3 \\ =-4-3 \\ =-7 \end{gathered}[/tex]And as it's not equal to zero, its not a solution.
b: x=1/3
In this case
[tex]\begin{gathered} 3(\frac{1}{3})^2+8(\frac{1}{3})-3=\frac{3}{9}+\frac{8}{3}-3 \\ =\frac{1}{3}+\frac{8}{3}-3 \\ =\frac{9}{3}-3 \\ =3-3=0 \end{gathered}[/tex]And as it's equal to zero, it's a solution.
c: x=1
In this case
[tex]\begin{gathered} 3(1)^2+8(3)-3=3+24-3 \\ =27-3 \\ =24 \end{gathered}[/tex]So, 1 is not a solution
d: x=3
In this case,
[tex]\begin{gathered} 3(3)^2+8(3)-3=3(9)+24-3 \\ =27+24-3 \\ =48 \end{gathered}[/tex]So 3 is not a solution.
e: x=-1/2
In this case,
[tex]\begin{gathered} 3(-\frac{1}{2})^2+8(-\frac{1}{2})-3=3(\frac{1}{4})-\frac{8}{2}-3 \\ =\frac{3}{4}-4-3 \\ =\frac{3}{4}-7 \\ =\frac{3}{4}-\frac{28}{7} \\ =-\frac{25}{7} \end{gathered}[/tex]So -1/2 is not a solution.
f: x=-3
In this case,
[tex]\begin{gathered} 3(-3)^2+8(-3)-3=3(9)-24-3 \\ =27-24-3 \\ =3-3 \\ =0 \end{gathered}[/tex]This means that 3 is a solution of the equation.
