completing the square

**emsypooh** · September 16th 2009, 08:37 AM

Hi everyone

could you please help me answer these two questions? Thanks! x

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.

**running-gag** · September 16th 2009, 08:52 AM

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 $x_2$ ?

**emsypooh** · September 16th 2009, 08:54 AM

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

**running-gag** · September 16th 2009, 09:00 AM

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

**emsypooh** · September 16th 2009, 09:06 AM

nopeee, we havent learnt any of this. we went from factorizing to this.

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

sorry about this..

**running-gag** · September 16th 2009, 09:22 AM

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

**emsypooh** · September 16th 2009, 09:26 AM

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

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

**running-gag** · September 16th 2009, 09:35 AM

Originally Posted by emsypooh

thank you so, so, so, so much!!!! i really, really appreciate it

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

This one is not finished !

**emsypooh** · September 16th 2009, 09:37 AM

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!

**running-gag** · September 16th 2009, 10:34 AM

No

I mean the second exercise is not yet finished

**emsypooh** · September 16th 2009, 10:58 AM

aahhhh okay, i understand

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