# completing the square

• September 16th 2009, 08:37 AM
emsypooh
completing the square
Hi everyone (Hi)

1. Show that there is no real value of the constant c for which the equation

cx^2 + (4c+1)x + (c+2) = 0

2. Given that the roots of the equation x^2+ax+(a+2) = 0 differ by 2, find the possible values of the constant a. Hence state the possible values of the roots of the equation.
• September 16th 2009, 08:52 AM
running-gag
Quote:

Originally Posted by emsypooh
2. Given that the roots of the equation x^2+ax+(a+2) = 0 differ by 2, find the possible values of the constant a. Hence state the possible values of the roots of the equation.

Hi

Let $x_1$ and $x_2$ the solutions of the equation x²+ax+(a+2) = 0

What relations can you write involving $x_1$ and $x_2$ ?
• September 16th 2009, 08:54 AM
emsypooh
Quote:

Originally Posted by running-gag
Hi

Let $x_1$ and $x_2$ the solutions of the equation x²+ax+(a+2) = 0

What relations can you write involving $x_1$ and $x_2$ ?

Hi there :)

I'm sorry, i sound so stupid, but i really don't know (Thinking)
• September 16th 2009, 09:00 AM
running-gag
The first one is of course $x_1 - x_2 = 2$ since you know that the roots differ by 2

Now you should also be able to write $x_1 + x_2 = ...$ and $x_1 \times x_2 = ...$

You should have learned this I suppose
• September 16th 2009, 09:06 AM
emsypooh
nopeee, we havent learnt any of this. we went from factorizing to this.

so the answers are x1+x2 and x1 (times) x2?

• September 16th 2009, 09:22 AM
running-gag
Okay
Since $x_1$ and $x_2$ are the roots of x²+ax+(a+2) you can write
$x^2+ax+(a+2) = (x-x_1)(x-x_2)$

Expanding the RHS $x^2+ax+(a+2) = x^2-(x_1+x_2)x + x_1 x_2$

Therefore
$x_1 + x_2 =-a$
and
$x_1 x_2 = a+2$

Now you know $x_1 + x_2 =-a$ and $x_1 - x_2 =2$

That will give you $x_1$ and $x_2$ with respect to a
• September 16th 2009, 09:26 AM
emsypooh
Quote:

Originally Posted by running-gag
Okay
Since $x_1$ and $x_2$ are the roots of x²+ax+(a+2) you can write
$x^2+ax+(a+2) = (x-x_1)(x-x_2)$

Expanding the RHS $x^2+ax+(a+2) = x^2-(x_1+x_2)x + x_1 x_2$

Therefore
$x_1 + x_2 =-a$
and
$x_1 x_2 = a+2$

Now you know $x_1 + x_2 =-a$ and $x_1 - x_2 =2$

That will give you $x_1$ and $x_2$ with respect to a

thank you so, so, so, so much!!!! i really, really appreciate it (Giggle)

do you know how to work out the other question too? x
• September 16th 2009, 09:35 AM
running-gag
Quote:

Originally Posted by emsypooh
thank you so, so, so, so much!!!! i really, really appreciate it (Giggle)

do you know how to work out the other question too? x

This one is not finished !
• September 16th 2009, 09:37 AM
emsypooh
Quote:

Originally Posted by running-gag
This one is not finished !

oh yeah whoops!

the full question is,

show that there is no real value of the constance c for which the equation

cx^2 + (4c+1)x + (c+2) = 0

has a repeated root.

whoops! :)
• September 16th 2009, 10:34 AM
running-gag
No

I mean the second exercise is not yet finished
• September 16th 2009, 10:58 AM
emsypooh
aahhhh okay, i understand :)