Respuesta :

Answer:

Option C) [tex]{\{0.5,0}\}[/tex] is correct

The solution to the given equation is [tex]{\{0.5,0}\}[/tex]

Step-by-step explanation:

Given quadratic equation is [tex]-2x^2+x=0[/tex]

To solve the equation by completing the square :

[tex]-2x^2+x=0[/tex]

[tex]-2(x^2-\frac{x}{2})=0[/tex]

[tex]x^2-\frac{x}{2}=0[/tex]

Rewritting the above equation as below :

[tex]x^2-2(x)\frac{1}{4}+\frac{1}{4^2}-\frac{1}{4^2}=0[/tex]

[tex](x-\frac{1}{4})^2-\frac{1}{4^2}=0[/tex]

[tex](x-\frac{1}{4})^2=\frac{1}{4^2}[/tex]

Taking square root on both sides we get

[tex]\sqrt{(x-\frac{1}{4})^2}=\sqrt{\frac{1}{4^2}}[/tex]

[tex](x-\frac{1}{4})=\pm\frac{1}{4}[/tex]

[tex]x=\pm\frac{1}{4}+\frac{1}{4}[/tex]

[tex]x=+\frac{1}{4}+\frac{1}{4}[/tex] and [tex]x=-\frac{1}{4}+\frac{1}{4}[/tex]

[tex]x=\frac{2}{4}[/tex] and [tex]x=0[/tex]

Therefore [tex]x=\frac{1}{2}[/tex] and x=0

[tex]x=0.5[/tex] and x=0

Therefore the solution to the given equation is [tex]{\{0.5,0}\}[/tex]

Option C) [tex]{\{0.5,0}\}[/tex] is correct

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