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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?

May 20, 2019

#1
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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.

May 20, 2019
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oohhhh...I see, wow, what shall I do without you guys??...thank you very much Rom, I really do appreciate...Have a blessed day..

juriemagic  May 20, 2019
#3
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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..

juriemagic  May 20, 2019
#4
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The square root of a negative number produces an imaginary number.

The problem specifies it wants the real roots.

Rom  May 20, 2019