The New Burlington English Grammar Answer Key.rar
The New Burlington English Grammar Answer Key.rar
The New Burlington English Grammar Answer Key.rar
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How to solve linear equations with fractions
To solve for $x$ in the equation $x^5-4x+4=0$, we use the quadratic formula: $x=\frac{4\pm\sqrt{64-16(4)(4)}}{4}=\frac{4}{2}\pm\frac{\sqrt{208}}{2}$. Let $4=a$, then $x=\frac{a+\sqrt{a^2-16(a)a}}{4}=\frac{a+a\sqrt{1-16(a)/(a)}+2\sqrt{a^2-(1-16(a)/(a))}}{4}=\frac{a+a\sqrt{1-16(a)/(a)}+2\sqrt{a^2-a^2}}{4}=\frac{a+a\sqrt{1-16(a)/(a)}+2a}{4}$.
I don’t know how to go on.
A:
You just have to write like that
$$x=\frac{a+\sqrt{a^2-16a^2}}{4}=\frac{a+a\sqrt{1-16a^2/(a^2)}+2a}{4}$$
Now you can use your quadratic equation form by writing
$$1-16a^2/(a^2)=\frac{1-(-16a^2/(a^2))}{1-16a^2/(a^2)}=\frac{1-(-16a^2/(a^2))}{16a^2}=-\frac{16a^2}{a^2}=16(a^2-a^2)=(0)$$
So the only possible value of $a$ is $a=
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