Answer to Question #289380 in Algebra for nekisha

Question #289380

find points of intersection of

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

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


1
Expert's answer
2022-01-27T01:00:16-0500

f(x)=x4+8x3+16x2f(x)=x^4+8x^3+16x^2

g(x)=54x216xg(x)=5-4x^2-16x

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

x4+8x3+16x2=54x216xx^4+8x^3+16x^2=5-4x^2-16x

x4+8x3+16x25+4x2+16x=0x^4+8x^3+16x^2-5+4x^2+16x=0

x4+8x3+16x2+4x2+16x5=0x^4+8x^3+16x^2+4x^2+16x-5=0

x4+8x3+20x2+16x5=0x^4+8x^3+20x^2+16x-5=0

x4+x^4+ 8x3+20x2+16x=58x^3+20x^2+16x=5

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

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

[x2+4x1][x2+4x+5]=0[x^2+4x-1][x^2+4x+5]=0


Hence, we set [x2+4x1]=0and[x2+4x+5]=0[x^2+4x-1]=0\hspace{0.2cm}and\hspace{0.2cm}[x^2+4x+5]=0

[x2+4x1]=0[x^2+4x-1]=0

solving quadratically, we have

    x=b±b24ac2a    x=4±424(1)(1)2(1)\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)}


    4±16+42\displaystyle \implies\frac{-4\pm\sqrt{16+4}}{2}


    4±202\displaystyle \implies\frac{-4\pm\sqrt{20}}{2}


    42±202\displaystyle \implies\frac{-4}{2}\pm\frac{\sqrt{20}}{2}


    2±252\displaystyle \implies-2\pm\frac{2\sqrt{5}}{2}


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


    2+5or    25\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

[x2+4x+5]=0[x^2+4x+5]=0

    x=4±424(1)(5)2(1)\displaystyle \implies x=\frac{-4\pm\sqrt{4^2-4(1)(5)}}{2(1)}


    4±16202\displaystyle \implies\frac{-4\pm\sqrt{16-20}}{2}


    4±42\displaystyle \implies\frac{-4\pm\sqrt{-4}}{2}


    42±42\displaystyle \implies\frac{-4}{2}\pm\frac{\sqrt{-4}}{2}


    2±2i2\displaystyle \implies-2\pm\frac{2i}{2}


    2±i\displaystyle \implies-2\pm i

    2+ior2i\implies -2+i\hspace{0.2cm}or\hspace{0.2cm}-2-i

Finding the roots of the polynomial,

x=(52,2+5,2i,2+i)x=(-\sqrt5-2,-2+\sqrt5,-2-i,-2+i)

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

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

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

y=54x216xy=5-4x^2-16x

    y=(1,25)\implies y=(1,25)

The points of intersection are

(52,1),(2+5,1),(2i,25),(2+i,25).(-\sqrt5-2,1),(-2+\sqrt5,1),(-2-i,25),(-2+i,25).




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment