I am working on a problem which requires the quadratic formula to solve. The problem is as follows:

For which value(s) of p does the equation px^2 - 10x + p = 0 have exactly two solutions?

I have plugged in the numbers and come up with:

x = 10 +/- (√-10^2 - 4p^2)/2p

From there I am unsure of what I need to do to get p by itself for the solution.

Any help would be appreciated.

Thanks,

Matty

2. Originally Posted by AnAmericanInNederlands25
I am working on a problem which requires the quadratic formula to solve. The problem is as follows:

For which value(s) of p does the equation px^2 - 10x + p = 0 have exactly two solutions?

I have plugged in the numbers and come up with:

x = 10 +/- (√-10^2 - 4p^2)/2p

From there I am unsure of what I need to do to get p by itself for the solution.

Any help would be appreciated.

Thanks,

Matty
You'd use the discriminant ($\displaystyle \Delta = b^2 - 4ac$) but I don't understand the question.

If $\displaystyle \Delta > 0$ then you have two, different real solutions
If $\displaystyle \Delta = 0$ then you have two, identical real solutions
If $\displaystyle \Delta < 0$ then you have two, different complex solutions

The way your question is worded is ambiguous since all three cases give two solutions.

If they meant equal solutions solve $\displaystyle b^2 - 4ac = 0$

$\displaystyle (-10)^2 - 4 \cdot p \cdot p = 0$

$\displaystyle 100-4p^2 = 0$

$\displaystyle (10-2p)(10+2p) = 0$

$\displaystyle p = \pm 5$