# Thread: Need Help with problems

1. ## Need Help with problems

I missed class and I am totally lost with these problems. What we have to do is find all real zeros of the function algebraically. My teacher had this equation up on the board but I can;t find it anywhere so any help will help
-bsqaurerooe+-4acdivided by 2a

f(x)=x^2-16

f(x)=t^2-4t^2+4t

g(t)=t^5-6t^3+9t

2. Originally Posted by jvrivera
I missed class and I am totally lost with these problems. What we have to do is find all real zeros of the function algebraically. My teacher had this equation up on the board but I can;t find it anywhere so any help will help
-bsqaurerooe+-4acdivided by 2a

f(x)=x^2-16

f(x)=t^2-4t^2+4t

g(t)=t^5-6t^3+9t
$x^2 - 16 = 0$

$x^2 = 16$

$x = \pm 4$.

$t^2 - 4t^2 + 4t = 0$

$4t - 3t^2 = 0$

$t(4 - 3t) = 0$

$t = 0$ or $4 - 3t = 0$

$t = 0$ or $t = \frac{4}{3}$.

$t^5-6t^3+9t = 0$

$t(t^4 - 6t^2 + 9) = 0$

$t(t^2 - 3)^2 = 0$

$t = 0$ or $t^2 - 3 = 0$

$t = 0$ or $t = \pm \sqrt{3}$.

3. Originally Posted by jvrivera
I missed class and I am totally lost with these problems. What we have to do is find all real zeros of the function algebraically. My teacher had this equation up on the board but I can;t find it anywhere so any help will help
-bsqaurerooe+-4acdivided by 2a
I think you mean the "quadratic formula" which is
"if $ax^2+ bx+ c= 0$, then $x= \frac{-b\pm \sqrt{b^2- 4ac}}{2a}$" (you forgot the $b^2$).

f(x)=x^2-16
As Prove It said, you can factor. If you want to use the quadratic formula, a= 1, b= 0, c= -16. It is always better to factor if you can. Save the quadratic formula for problems where you can't factor.

f(x)=t^2-4t^2+4t
This is probably NOT what you intend. This is $-3t^2+ 4t= t(-3t+ 4)= 0$ which has roots t= 0 and t= 4/3. But I suspect you meant $t^2- 4t+ 4= (t- 2)^2= 0$ which has the root t= 2.
Again, you could use the quadratic formula with a= 1, b= -4, c= 4.

g(t)=t^5-6t^3+9t
You can factor this by taking out a single t, $t(t^4- 6t^2+ 9)= 0$ which has t= 0 as a root. Now let $x= t^2$ and the other factor becomes $x^2- 6x+ 9= (x- 3)^2= 0$. That has the single root x= 3 so $t^2= 3$ and then $t= \pm\sqrt{3}$