+1124 Hi all,
The roots of a quadratic equation are: \(x=5 \pm \sqrt{2p-1}\)
calculate the value of p for which the roots are real....
so it's: \(0=5 + \sqrt{2p-1}\)
\( \sqrt{2p-1} = -5\)
\(2p-1=-5^2\)
\(2p=26\)
\(p=13\)
and if I go the other way:
\(0=5 - \sqrt{2p-1}\)
\( \sqrt{2p-1} = +5\)
it also ends up being 13.
so I believe there is only one value for p..am I right?
This next question's got me...it says to calculate ONE value for p for which the roots are equal?..
and then asks for one value for which the roots are un-equal and rasional?
please help me??..
+6251 yikes, no. You are told those are the roots of some unspecified quadratic equation.
In order for them to be real we need the argument to that square root to be non-negative.
Otherwise that bit would contribute an imaginary part.
Ok so that means
\(2p-1\geq 0\\ 2p\geq 1\\ p \geq \dfrac 1 2\)
so there's not a single value of p that causes the roots to be real but a half closed interval of them.
For two equal valued roots we simply have the bit in the square root be equal to 0.
\(2p-1=0\\ p = \dfrac 1 2\)
For rational but unequal roots the bit in the square root has to be a perfect square.
\(2p-1 = k^2,~k \in \mathbb{N}\\ p = \dfrac{k^2+1}{2},~k \in \mathbb{N}\)
for example p = 5 provides 2 rational but unequal roots.
+1124 oohhhh...I see, wow, what shall I do without you guys??...thank you very much Rom, I really do appreciate...Have a blessed day..
+1124 Rom just a question please,
Real numbers include negative numbers as well not so?...so then, why do we have to have the argument to that square root, to be non-negative?..sorry, I just do not quite understand?..thanx for your time..
