# [SOLVED] help

• Mar 23rd 2009, 05:15 AM
princess_21
[SOLVED] help
If I can't factor a quadratic equation what will I do??
$x^2+2x+7\ge 0$
I don't know the latex for this.. thanks. I tried extracting the roots but it turned out as square root of a negative number?? Am I in the right track?
• Mar 23rd 2009, 05:30 AM
skeeter
Quote:

Originally Posted by princess_21
If I can't factor a quadratic equation what will I do??
$x^2+2x+7 > or = 0$
I don't know the latex for this.. thanks. I tried extracting the roots but it turned out as square root of a negative number?? Am I in the right track?

since $b^2-4ac < 0$ , then the quadratic has no real roots ... which tells you something very valuable regarding its graph.
• Mar 23rd 2009, 05:32 AM
princess_21
Quote:

Originally Posted by skeeter
since $b^2-4ac < 0$ , then the quadratic has no real roots ... which tells you something very valuable regarding its graph.

then what will be the answer in this question?? (Worried) thanks I dont know where to start. can you guide on where to start the process? and then I will try doing the rest thanks
• Mar 23rd 2009, 05:44 AM
SirNostalgic
Quote:

Originally Posted by skeeter
since $b^2-4ac < 0$ , then the quadratic has no real roots ... which tells you something very valuable regarding its graph.

This means theres no intersection with the x-axis. if you simply need to sketch the graph find the turning point of the quadratic $\frac{d(y)}{d(x)} = 0$ should give you the X-coordinate of the turning point. Put that X-coordinate value into the non-differentiated equation and you'll get your Y-coordinate of the turning point
• Mar 23rd 2009, 05:49 AM
princess_21
Quote:

Originally Posted by SirNostalgic
This means theres no intersection with the x-axis. if you simply need to sketch the graph find the turning point of the quadratic $\frac{d(y)}{d(x)} = 0$ should give you the X-coordinate of the turning point. Put that X-coordinate value into the non-differentiated equation and you'll get your Y-coordinate of the turning point

sorry I didn't understand what you said. can you explain further for me? thanks for your time (Thinking)
• Mar 23rd 2009, 05:50 AM
SirNostalgic
$x^2+2x+7$

$\frac{d(y)}{d(x)} = 2x+2$
$\frac{d(y)}{d(x)} = 0$

$2x+2 = 0$
$2x = -2$
$x = -1$

x = -1 is the x-coordinate of the Turning point, plug into the original equation for the Y-value of this Turning point
$x^2+2x+7 (x= -1)
$

$(-1)^2 +2(-1) +7$
= 6

Coordinates of Turning point (-1,6) I'm guessing this is a minimum :p

i swear to god my computer pasted what i wrote twice. What in gods name happened there lol
• Mar 23rd 2009, 05:59 AM
stapel
Quote:

Originally Posted by princess_21
then what will be the answer in this question?? (Worried) thanks I dont know where to start. can you guide on where to start the process? and then I will try doing the rest thanks

Graph the quadratic. Look at the picture. Look at the inequality.

You are asked to find the x-values for which the quadratic is at or above the x-axis. What does the picture show you?

(Wink)
• Mar 23rd 2009, 06:06 AM
princess_21
Quote:

Originally Posted by stapel
Graph the quadratic. Look at the picture. Look at the inequality.

You are asked to find the x-values for which the quadratic is at or above the x-axis. What does the picture show you?

(Wink)

i still can't understand. as you see I learn from examples and I can't understand from explanations. can you give an example? thanks
• Mar 23rd 2009, 06:31 AM
stapel
Quote:

Originally Posted by princess_21
i still can't understand. as you see I learn from examples and I can't understand from explanations. can you give an example? thanks

If the various worked examples in the lesson (provided in the link) were not sufficiently helpful, then try studying a few more online lessons. (I'm assuming you understand the x,y-plane and how to graph linear equations.)

Once you've learned how to do the graph, draw the picture. (Wink)
• Mar 23rd 2009, 06:46 AM
Quote:

Originally Posted by princess_21
If I can't factor a quadratic equation what will I do??
$x^2+2x+7 > or = 0$
I don't know the latex for this.. thanks. I tried extracting the roots but it turned out as square root of a negative number?? Am I in the right track?

A simple answer to your question is

-Roots are not real as square root of negative is not real

-Meaning graph never touches x axis , or in other words the quadratic function never has zero value

-This means that the quadratic is either always positive or always negative
(since to change from +ve to -ve you need to cross 0 )

Result:
Value of equation for all x is either always positive or always negative

----------------------------
Application :

Put x = any number , lets take it as 2

$2^2+2\times 2+7 = 15$

15>0, as derived earlier for all values of x this is equality holds that

$x^2+2x+7 > 0$

thus answer is every real x
-------------------------------------------

No real need to draw a graph or learning it for answering this
• Mar 23rd 2009, 07:07 AM
princess_21
Quote:

Originally Posted by ADARSH
A simple answer to your question is

-Roots are not real as square root of negative is not real

-Meaning graph never touches x axis , or in other words the quadratic function never has zero value

-This means that the quadratic is either always positive or always negative
(since to change from +ve to -ve you need to cross 0 )

Result:
Value of equation for all x is either always positive or always negative

----------------------------
Application :

Put x = any number , lets take it as 2

$2^2+2\times 2+7 = 15$

15>0, as derived earlier for all values of x this is equality holds that

$x^2+2x+7 > 0$

thus answer is every real x
-------------------------------------------

No real need to draw a graph or learning it for answering this

so my answer will be all real numbers??
• Mar 23rd 2009, 07:10 AM
Quote:

Originally Posted by princess_21
so my answer will be all real numbers??

Yes (Yes) ! (Itwasntme)

You are very old now (Tongueout) :p ..:D
• Mar 23rd 2009, 07:13 AM
princess_21
Quote:

Originally Posted by ADARSH
Yes (Yes) ! (Itwasntme)

You are very old now (Tongueout) :p ..:D

thanks. (Crying) i need to do better in my exam.. i need to pass after failing in the first one.

***atleast you're older
• Mar 23rd 2009, 07:14 AM
princess_21