Q about roots of quadratic eqn: Given a, b, c are lengths of sides of triangle...

lai001

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Hi. Recently my friend ask me a question:
Given that a, b and c is length of sides of a triangle and a>b,is the quadratic equation cx^2 +((a^2-b^2)^(1/2))x +(2a-c)/4=0 has no real solution?
I completely do not know where to start?
Maybe using b^2-4ac<0?And then I completely had no idea.Can somebody help me please?Thanks.
 
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Hi. Recently my friend ask me a question:
Given that a, b and c is sides of a triangle and a>b,is the quadratic equation cx^2 +((a^2-b^2)^(1/2))x +(2a-c)/4=0 has no real solution?
I completely do not know where to start?
Maybe using b^2-4ac<0?And then I completely had no idea.Can somebody help me please?Thanks.
Why is your friend asking you this question - and you to us?

Please let him/her ask the question directly - including some work
 
Why is your friend asking you this question - and you to us?

Please let him/her ask the question directly - including some work
My friend asked me personally, I cannot solve it, so I ask here, is there any part of the question you can't understand,here is my approach to this question :
Using b^2-4ac,
a^2-b^2-4(c)(2a-c/4)<0
a^2-b^2-2ac+c^2<0
c^2-2ac+a^2-b^2<0
It's here that I am stuck...
 
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My friend asked me personally, I cannot solve it, so I ask here, is there any part of the question you can't understand,here is my approach to this question :
Using b^2-4ac,
a^2-b^2-4(c)(2a-c/4)<0
a^2-b^2-2ac+c^2<0
c^2-2ac+a^2-b^2<0
It's here that I am stuck...
Well ... we.....I do not work through third party very well....
Hint: use difference of squares formula...
 
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My friend asked me personally, I cannot solve it, so I ask here, is there any part of the question you can't understand,here is my approach to this question :
Using b^2-4ac,
a^2-b^2-4(c)(2a-c/4)<0
a^2-b^2-2ac+c^2<0
c^2-2ac+a^2-b^2<0
It's here that I am stuck...
Continuing from your last line, we can factor to get

\(\displaystyle c^2 - 2ac + a^2 - b^2 = (a - c)^2 - b^2 = (a-c-b)(a-c+b).\)

So the equation has real solutions if and only if \(\displaystyle (a-c-b)(a-c+b)\geq0\). Since \(\displaystyle a\), \(\displaystyle b\), and \(\displaystyle c\) are lengths of sides in a triangle, we must have \(\displaystyle a + b > c\) so the second factor must be bigger than zero. So the equation has solutions if and only if \(\displaystyle a - c - b\geq0\) if and only if \(\displaystyle a\geq b + c\). Hopefully you can see that this can't happen.
 
Your strategy seems fine to me. I'd recommend using different terminology for clarity's sake though. Since a, b, and c are the sides of the triangle, let's use \(\displaystyle \alpha, \beta, \text{ and } \gamma\) as the coefficients of the polynomial. Then, to test if the polynomial has no real solutions, we need to check that:

\(\displaystyle \beta^2 - 4\alpha\gamma < 0\)

Making the appropriate substitutions:

\(\displaystyle a^2 - b^2 - 4(c)\left(\dfrac{2a-c}{4}\right) < 0\)

Simplifying and rearranging terms:

\(\displaystyle a^2 - b^2 + c^2 - 2ac < 0\)

This is exactly what you ended up with, so we're good on that front. My next step would be to use the Law of Cosineshttps://www.mathsisfun.com/algebra/trig-cosine-law.html to make a substitution for b2. Where does that lead you? I'd note that if the product of two terms is less than 0, then we know that exactly one of them must be negative. That will help you complete the proof.
 
Your strategy seems fine to me. I'd recommend using different terminology for clarity's sake though. Since a, b, and c are the sides of the triangle, let's use \(\displaystyle \alpha, \beta, \text{ and } \gamma\) as the coefficients of the polynomial. Then, to test if the polynomial has no real solutions, we need to check that:

\(\displaystyle \beta^2 - 4\alpha\gamma < 0\)

Making the appropriate substitutions:

\(\displaystyle a^2 - b^2 - 4(c)\left(\dfrac{2a-c}{4}\right) < 0\)

Simplifying and rearranging terms:

\(\displaystyle a^2 - b^2 + c^2 - 2ac < 0\)

This is exactly what you ended up with, so we're good on that front. My next step would be to use the Law of Cosines to make a substitution for b2. Where does that lead you? I'd note that if the product of two terms is less than 0, then we know that exactly one of them must be negative. That will help you complete the proof.

It is actually much simpler than that:

\(\displaystyle a^2 - b^2 + c^2 - 2ac < 0\)

\(\displaystyle (a-c)^2 - b^2 \)

\(\displaystyle [a + b - c][a - (c+b)] \)

Since a, b & c are sides of an Euclidian triangle - the first factor is always positive and the second factor is always negative.

So the product is always negative → <0

In previous post, the OP referred to Lambert function - I was surprised that s/he did not see this reduction!!
 
Continuing from your last line, we can factor to get

\(\displaystyle c^2 - 2ac + a^2 - b^2 = (a - c)^2 - b^2 = (a-c-b)(a-c+b).\)

So the equation has real solutions if and only if \(\displaystyle (a-c-b)(a-c+b)\geq0\). Since \(\displaystyle a\), \(\displaystyle b\), and \(\displaystyle c\) are lengths of sides in a triangle, we must have \(\displaystyle a + b > c\) so the second factor must be bigger than zero. So the equation has solutions if and only if \(\displaystyle a - c - b\geq0\) if and only if \(\displaystyle a\geq b + c\). Hopefully you can see that this can't happen.
Why must a+b>c?
 
It is actually much simpler than that:

\(\displaystyle a^2 - b^2 + c^2 - 2ac < 0\)

\(\displaystyle (a-c)^2 - b^2 \)

\(\displaystyle [a + b - c][a - (c+b)] \)

Since a, b & c are sides of an Euclidian triangle - the first factor is always positive and the second factor is always negative.

So the product is always negative → <0

In previous post, the OP referred to Lambert function - I was surprised that s/he did not see this reduction!!
Please don't overestimate me, something like Lambert function can be found in the internet....
 
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Look buddy, this is getting ridiculous...

Please read up on triangle inequality:
https://en.wikipedia.org/wiki/Triangle_inequality

...then come back and ask your question clearly,
and in proper English: then we'll be able to help :cool:
Ok, I know how to solve this question now... Sorry for my improper English...And also can you tell me where my English in improper in my question?This will help me to improve my English skills. Thanks
 
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