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## Need help solving weirdly phrased quadratic equations

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
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### Need help solving weirdly phrased quadratic equations

by algebraid on Thu Jul 02, 2009 6:57 pm

Since this site has already helped me a lot I thought I would seek your help with a quadratic equation that I can't wrap my head around.

The problem is phrased as follow (but I am translating from dutch):
In the quadratic equation 2x^2 - 3x + p = 0 one of the answers is equal to 2. Calculate p and the other answer.

Is it just me or is this pretty weirdly phrased?

So I tried solving it with the quadratic formula(x), but there is no answer that equals 2.(or is there?) And I have no idea how to calculate p.

The final answer using the quadratic formula is:

$\LARGE x=\frac{3\pm\sqrt{-8p+9}}{4}$

If anyone could shed some light on this I would be very grateful.
algebraid

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by stapel_eliz on Thu Jul 02, 2009 8:01 pm

algebraid wrote:In the quadratic equation 2x^2 - 3x + p = 0 one of the answers is equal to 2. Calculate p and the other answer.

The Quadratic Formula is probably the best way to go. It gives you:

. . . . .$x\, =\, \frac{3\, \pm\, \sqrt{9\, -\, 8p}}{4}$

For this to come out "even", which a nice whole-number value for the solution, the value of 9 - 8p needs to be a perfect square. The squares are 0, 1, 4, 9, 16, 25, 36, etc. Let's see if we can narrow the list down a bit by applying what we know about solving radical equations....

. . . . .$\frac{3\, \pm\, \sqrt{9\, -\, 8p}}{4}\, =\, 2$

. . . . .$3\, \pm\, \sqrt{9\, -\, 8p}\, =\, 8$

. . . . .$\pm\sqrt{9\, -\, 8p}\, =\, 5$

. . . . .$9\, -\, 8p\, =\, 25$

. . . . .$-16\, =\, 8p$

What then is the value of "p"? Plugging this into the Quadratic Formula, what is the other solution?

stapel_eliz

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Joined: Mon Dec 08, 2008 4:22 pm

### Re: Need help solving weirdly phrased quadratic equations

by algebraid on Fri Jul 03, 2009 10:00 am

First, thank you very much for your help.

Second, I think the answer to the question is that the solution that is equal to 2 is p = -2. The other answer would be p=0. So p=0,-2. In the original equation p = 2x^2 +3x, so it makes sense to do x first and go from there to calculate p.

Can you tell me if this is correct?
algebraid

Posts: 3
Joined: Thu Jul 02, 2009 6:32 pm

by stapel_eliz on Fri Jul 03, 2009 12:53 pm

If p = -2, then the quadratic equation becomes 2x2 - 3x - 2 = 0. If x = 0, then the equation is -2 = 0, which is not true. So the other solution cannot be "x = 0".

You are given that one solution is "x = 2". This was used to find the value of p. Now that you have found the (one and only) value of p, you need to factor the quadratic, or else use the Quadratic Formula, to confirm the one (given) solution and find the other (requested) solution.

stapel_eliz

Posts: 1701
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Need help solving weirdly phrased quadratic equations

by algebraid on Fri Jul 03, 2009 8:20 pm

So if I have this correct:
p = -2 x = -0.5, 2

Again thank you so much for your help. I'm really glad I've found you to help me with this, because I'm doing a self-study for an upcoming entrance exam (important!) and I don't have anyone else I can ask to help me with this. Maybe I will speak/write to you again.
algebraid

Posts: 3
Joined: Thu Jul 02, 2009 6:32 pm

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