Respuesta :

Answer:

The equation is: [tex]x^2-x+\frac{1}{2}[/tex]

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

(B)[tex]5x^2-5x+\frac{5}{2}=0[/tex]

(C) [tex]-\frac{1}{2}[/tex]

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

d) √3

[tex](\sqrt{3}) x^2-x\sqrt{3}+\frac{\sqrt{3}}{2}=0[/tex]

Step-by-step explanation:

If the roots of the quadratic equation are: [tex]\frac{\sqrt{3}+1}{2} and \frac{\sqrt{3}-1}{2}[/tex]

This means:

x=[tex]\frac{\sqrt{3}+1}{2}[/tex] OR [tex]x= \frac{\sqrt{3}-1}{2}[/tex]

x-[tex]x-\frac{\sqrt{3}+1}{2}=0 \: or\: x- \frac{\sqrt{3}-1}{2}=0[/tex]=0 or x-

If a=0 or b=0, then ab=0

Therefore:

[tex](x-\frac{\sqrt{3}+1}{2})(x-\frac{\sqrt{3}-1}{2})=0\\x^2-\frac{x(\sqrt{3}-1)}{2}-\frac{x(\sqrt{3}+1)}{2}+(\frac{\sqrt{3}-1)}{2})(\frac{\sqrt{3}+1}{2})=0\\x^2-x+\frac{1}{2}=0[/tex]

Therefore the coefficient of [tex]x^2[/tex]  here is 1.

b) To make 5 the coefficient of [tex]x^2[/tex]

Simply multiply [tex]x^2-x+\frac{1}{2}=0[/tex] by 5

This gives: [tex]5x^2-5x+\frac{5}{2}=0[/tex]

c) [tex]-\frac{1}{2}[/tex]

To make [tex]-\frac{1}{2}[/tex] the coefficient of [tex]x^2[/tex]

Simply multiply [tex]x^2-x+\frac{1}{2}=0[/tex] by [tex]-\frac{1}{2}[/tex]

This gives: [tex]-\frac{1}{2}x^2+\frac{1}{2}x-\frac{1}{4}=0[/tex]

d) √3

To make  [tex]\sqrt{3}[/tex] the coefficient of [tex]x^2[/tex]

Simply multiply [tex]x^2-x+\frac{1}{2}=0[/tex] by [tex]\sqrt{3}[/tex]

This gives: [tex](\sqrt{3}) x^2-x\sqrt{3}+\frac{\sqrt{3}}{2}=0[/tex]

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