Question

find points of intersection of

f(x)=x^4+8x^3+16x^2

g(x)=5-4x^2-16x

Answer 1

$f(x)=x^4+8x^3+16x^2$

$g(x)=5-4x^2-16x$

To find the point of intersection, let's equate both equations. This gives

$x^4+8x^3+16x^2=5-4x^2-16x$

$x^4+8x^3+16x^2-5+4x^2+16x=0$

$x^4+8x^3+16x^2+4x^2+16x-5=0$

$x^4+8x^3+20x^2+16x-5=0$

$x^4+$ $8x^3+20x^2+16x=5$

$x(x+2)^2(x+4)=5$

$\left[x(x+4)-1\right]\left[x(x+4)+5\right]=0$

$[x^2+4x-1][x^2+4x+5]=0$

Hence, we set $[x^2+4x-1]=0\hspace{0.2cm}and\hspace{0.2cm}[x^2+4x+5]=0$

$[x^2+4x-1]=0$

solving quadratically, we have

$\displaystyle \implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\displaystyle \implies x=\frac{-4\pm\sqrt{4^2-4(1)(-1)}}{2(1)}$

$\displaystyle \implies\frac{-4\pm\sqrt{16+4}}{2}$

$\displaystyle \implies\frac{-4\pm\sqrt{20}}{2}$

$\displaystyle \implies\frac{-4}{2}\pm\frac{\sqrt{20}}{2}$

$\displaystyle \implies-2\pm\frac{2\sqrt{5}}{2}$

$\displaystyle \implies-2\pm{\sqrt{5}}$

$\displaystyle \implies-2+{\sqrt{5}}\hspace{0.2cm}or\hspace{0.2cm}\displaystyle \implies-2-{\sqrt{5}}$

Also, solving the second Quadratic equation similarly, we have

$[x^2+4x+5]=0$

$\displaystyle \implies x=\frac{-4\pm\sqrt{4^2-4(1)(5)}}{2(1)}$

$\displaystyle \implies\frac{-4\pm\sqrt{16-20}}{2}$

$\displaystyle \implies\frac{-4\pm\sqrt{-4}}{2}$

$\displaystyle \implies\frac{-4}{2}\pm\frac{\sqrt{-4}}{2}$

$\displaystyle \implies-2\pm\frac{2i}{2}$

$\displaystyle \implies-2\pm i$

$\implies -2+i\hspace{0.2cm}or\hspace{0.2cm}-2-i$

Finding the roots of the polynomial,

$x=(-\sqrt5-2,-2+\sqrt5,-2-i,-2+i)$

let $y=f(x)=g(x)$

substituting $x$ into either $f(x) or g(x)$ we get the same value for $y.$

Let's substitute $x$ into $g(x)$

$y=5-4x^2-16x$

$\implies y=(1,25)$

The points of intersection are

$(-\sqrt5-2,1),(-2+\sqrt5,1),(-2-i,25),(-2+i,25).$