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]