Question

Someone please help. Grade 8 math I give brainlist. If answer is wrong then I will report. Extra points. Someone please help quickly. Ill give you a thanks as well

Answer 1

$\huge{\text{Solution}}$

Note that $f(x)=(x+1)(x^4-3x^2-9)$.

Zero product property.

$f(x)=0$ implies $x+1=0$ or $x^4-3x^2-9=0$.

$x+1=0$ implies $x=-1$.

Quadratic formula.

If we use the quadratic formula on $x^4-3x^2-9=0$,

$\implies x^2=\dfrac{3\pm3\sqrt{5} }{2}$

$\implies x=\pm\sqrt{\dfrac{3\pm3\sqrt{5} }{2}}$

Real solutions.

For $x=\pm\sqrt{\dfrac{3+3\sqrt{5} }{2}}$ there are two real solutions.

Imaginary solutions.

For $x=\pm\sqrt{\dfrac{3-3\sqrt{5} }{2}}$ there are two imaginary solutions, meaning they do not appear on the real plane.

Wavy curve method.

Substituting some values between solutions, we can find whether the section is positive or negative.

• $f(-\sqrt{\dfrac{3+3\sqrt{5} }{2}})=0$
• $f(-2)=(-2+1)(1-3-9)=11>0$
• $f(-1)=0$
• $f(0)=(0+1)(1-3-9)=-11<0$
• $f(\sqrt{\dfrac{3+3\sqrt{5} }{2}})=0$
• $f(3)=(3+1)(81-27-9)=180>0$

Conclusion.

By wavy curve method, the solution to $f(x)>0$ is $-\sqrt{\dfrac{3+3\sqrt{5} }{2}}<x<-1,x>\sqrt{\dfrac{3+3\sqrt{5} }{2}}$.

By wavy curve method, the solution to $f(x)<0$ is $x<-\sqrt{\dfrac{3+3\sqrt{5} }{2}},1<x<\sqrt{\dfrac{3+3\sqrt{5} }{2}}$.