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avatar+532 

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??..

 May 20, 2019
 #1
avatar+6043 
+1

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
 #2
avatar+532 
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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
avatar+532 
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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..

juriemagic  May 20, 2019
 #4
avatar+6043 
0

The square root of a negative number produces an imaginary number.

 

The problem specifies it wants the real roots.

Rom  May 20, 2019

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