A convenient way to find the zeros of \(f(x)=x^4-3x^2 +2\) is by factoring.
a) The equation,
\(x^4-3x^2 +2=0\)
can be rewritten as,
\((x^2)^2-3(x^2) +2=0\)
We can think of this equation as a quadratic equation in \(x^2\), with \(a=1,b=-3\:\: and \:\: c=2\).
Observe that \(ac=1 \times 2=2\).
We find two factors of 2 that adds up to \(-3\). These are, \(-2,-1\).
Now let us split the middle term. to obtain;
\((x^2)^2-(x^2) -2(x^2)+2=0\)
We can factor to get,
\((x^2)(x^2-1)-2((x^2-1)=0\)
We factor further to obtain;
\((x^2-1)((x^2-2)=0\)
\(\Rightarrow (x-1)(x+1((x- \sqrt{2})(x+ \sqrt{2})=0\)
Hence the zeroes of \(f(x)\) are;
\(x=1,x=-1,x= \sqrt{2},x=- \sqrt{2}\)
b) To find the line tangent, we must first, find the slope using differentiation. That is,
\(Slope\:\: function=f'(x)=4x^3-6x\)
At \(x=1\),
\(Slope=f'(1)=4(1)^3-6(1)=-2\)
Also, we need to determine the \(y\) value at \(x=1\). That is;
\(f(1)=(1)^4-3(1)^2+2=0\)
Now we can use the slope \(m=-2\) and the point \((1,0)\) to write ythe equation of the line tangent.
\(y-y_1 =m(x-x_1)\)
\(\Rightarrow y-0 =-2(x-1)\)
\(\Rightarrow y=-2x+2\)
c)
If the line tangent is parallel to the line \(y=-2x+4\), then
\(f'(x)=-2\)
Since parallel lines have the same slope.
\(\Righarrow 4x^3-6x=-2\)
\(\Righarrow 4x^3-6x+2=0\)
\(\Righarrow (x-1)(2x- ( \sqrt{3}-1)(2x+ ( \sqrt{3}-1)=0\)
Hence the x-coordinates are,
\(x=1,x= \frac{ \sqrt{3}-1}{2},x= -\frac{\sqrt{3}-1}{2}\)
