# Need help with what is probably a simple quadratic equation

• Nov 21st 2006, 11:01 AM
finlaymaguire
Need help with what is probably a simple quadratic equation
Hi, I really need a hand with the following question

Find the values for k for which the quadratic equation x^2-2x+21=2k(x-7) has equal roots

I don't really know what to do, I know the discriminant must equal 0 but actually putting that usefully is completely beyond me somehow.

Any help you could provide would be much appreciated.
• Nov 21st 2006, 11:29 AM
The Pondermatic
Solve for x in terms of k.

Get the equation in the form ax^2 + bx + c = 0. k will be in there somewhere in a,b, and/or c. Then, solve the quadratic. Instead of simplifying the qudratic formula to a constant, simplify it to terms with k.

Does that make sense? Shout at me if it doesn't.
• Nov 21st 2006, 11:31 AM
CaptainBlack
Quote:

Originally Posted by finlaymaguire
Hi, I really need a hand with the following question

Find the values for k for which the quadratic equation x^2-2x+21=2k(x-7) has equal roots

I don't really know what to do, I know the discriminant must equal 0 but actually putting that usefully is completely beyond me somehow.

Any help you could provide would be much appreciated.

rearrange the equation as:

$\displaystyle x^2-2(1+k)x+28=0$

Now the quadratic formula tells us that:

$\displaystyle x=\frac{2(1+k) \pm \sqrt{4(1+k)^2-4\times 28}}{2}$

There are two distinct roots unless what is under the root (the
discriminant is zero), when there is only one root (or two equal roots).
So the condition for a double root is that:

$\displaystyle 4(1+k)^2-4\times 28=0$

RonL
• Nov 21st 2006, 11:38 AM
finlaymaguire
ahhhh, I can't believe I couldn't see that, so would that make the values of k be 2 and -10/9?

Thanks for the help by the way, I really appreciate it.
• Nov 21st 2006, 11:56 AM
CaptainBlack
Quote:

Originally Posted by finlaymaguire
ahhhh, I can't believe I couldn't see that, so would that make the values of k2 and -10/9?

Thanks for the help by the way, I really appreciate it.

No it would be:

$\displaystyle k=-1 \pm \sqrt{28}$

RonL