Hi, I've got a review sheet due Monday in algebra and I really need some help with it!

Use quadratic formula to solve:
0=x(squared)-5x+6
0=x(squared)+x-10

Estimate the roots of the equation:
y=x(squared)-2x-8
y=x(squared)-4

Graph:

y>or equal to x(squared)
y<x(squared)-3

Hope I made that clear enough! Thanks!!!

2. Originally Posted by Lukeh
Hi, I've got a review sheet due Monday in algebra and I really need some help with it!

Use quadratic formula to solve:
0=x(squared)-5x+6
0=x(squared)+x-10

Estimate the roots of the equation:
y=x(squared)-2x-8
y=x(squared)-4

Graph:

y>or equal to x(squared)
y<x(squared)-3

Hope I made that clear enough! Thanks!!!
without LaTex, we write x squared as x^2

x^2-5x+6 = 0
you don't need the quadratic formula for this, but since you asked.

remember the quadratic formula.
for a quadratic ax^2 + bx + c = 0, x is given by:

x = [-b +/- sqrt(b^2 - 4ac)]/2a

x^2-5x+6 = 0
=> x = [5 +/- sqrt(25 - 4(6)]/2
=> x = [5 +/- sqrt(1)]/2
=> x = (5 + 1)/2 or (5 - 1)/2

so x = 3 , x = 2

x^2+x-10 = 0
x = [-1 +/- sqrt(1 - 4(-10))]/2
=> x = [-1 +/- sqrt(41)]/2
=> x = [-1 + sqrt(41)]/2 or x = [-1 - sqrt(41)]/2

y=x^2-2x-8
for the roots, we want x^2 - 2x - 8 = 0
=> (x - 4)(x + 2) = 0
=> x = 4 or x = -2 ...........these are the roots

y=x^2-4
this is the difference of two squares

we have
y = x^2 - 2^2
=> y = (x + 2)(x - 2)
for roots, we want
(x + 2)(x - 2) = 0

so x = -2 or x = 2

graph y>or equal to x^2
i'll get back to you on this

graph y<x^2-3
i'll get back to you on this as well

3. Originally Posted by Lukeh
Graph:

y>or equal to x(squared)
y<x(squared)-3
the shaded regions are what you want

the first graph is y >= x^2, in this graph, the line is included

the second graph is y < x^2 - 3, in this graph, the line is not included so you should draw a broken line.

4. Thanks a ton.