+0  
 
+8
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avatar+238 

In the triangle ABC, a,b,c are the 3 sides of the triangle.if a(1-x^2)+2bx+c(1+x^2)=0 is a qudratic equation. The equation have two same soloution and 3c=a+3b(a si the opposite side of angle A, b si the opposite side of angle B,c is the opposite side of angle C)

(1) What is the shape of the triangle?

(2)sinA+sinB=?

quinn  Sep 26, 2014

Best Answer 

 #3
avatar+19653 
+13

$$\small{\text{
$x^2 = 0 \Rightarrow c-a=0 \Rightarrow (2a-3b)= 0 \Rightarrow b=\frac{2}{3}a $ and $ c=\frac{a+3b}{3} \Rightarrow c=a $
}}$$

The triangle is isosceles triangle:

$$\small{\text{
$
\sin{(\frac{1}{2}(180\ensurement{^{\circ}} - 2\alpha))}=
\frac{\frac{2}{3}a}{a}=\frac{2}{3}=\sin{(90\ensurement{^{\circ}}-\alpha)}=\cos{(\alpha)} \quad \boxed{\cos{(\alpha)}=\frac{2}{3}}
$
}}
\\
\small{\text{
$
(\frac{2}{3}a)^2=2a^2-2a^2\cos{(\beta)} \Rightarrow \boxed{\cos{(\beta)}=\frac{7}{9}}
$
}}$$

$$\small{\text{
$
\sin{(\alpha)}=\sqrt{1-\cos^2{(\alpha)}}=\frac{1}{3}\sqrt{5} \qquad
\sin{(\beta)}=\sqrt{1-\cos^2{(\beta)}}= \frac{4}{9}\sqrt{2}
$
}}\\
\small{\text{
$
\sin{(\alpha)} + \sin{(\beta)}= \frac{1}{3}\sqrt{5} +\frac{4}{9}\sqrt{2}
$
}}$$

heureka  Sep 26, 2014
 #1
avatar+8258 
+8

Melody and CPHILL, here comes your favorite question!

DragonSlayer554  Sep 26, 2014
 #2
avatar+92806 
+13

Hi Quinn,

I am just looking at your question now and it has got me rather stumped.

Usin the quadratic formula and the discriminant=0  I get to

$$a^2+b^2-c^2=2(a-c)$$

$$c^2=a^2+b^2-2(a-c)$$

This looks similar to the cosine rule so I could say that

$$a-c=abcosC$$         [and we also know that    a=3c-3b]

Now  ummmm

This isn't helping.   

Have you got any thoughts Quinn?

Quinn must have just left -  Has anyone got any thoughts?

Melody  Sep 26, 2014
 #3
avatar+19653 
+13
Best Answer

$$\small{\text{
$x^2 = 0 \Rightarrow c-a=0 \Rightarrow (2a-3b)= 0 \Rightarrow b=\frac{2}{3}a $ and $ c=\frac{a+3b}{3} \Rightarrow c=a $
}}$$

The triangle is isosceles triangle:

$$\small{\text{
$
\sin{(\frac{1}{2}(180\ensurement{^{\circ}} - 2\alpha))}=
\frac{\frac{2}{3}a}{a}=\frac{2}{3}=\sin{(90\ensurement{^{\circ}}-\alpha)}=\cos{(\alpha)} \quad \boxed{\cos{(\alpha)}=\frac{2}{3}}
$
}}
\\
\small{\text{
$
(\frac{2}{3}a)^2=2a^2-2a^2\cos{(\beta)} \Rightarrow \boxed{\cos{(\beta)}=\frac{7}{9}}
$
}}$$

$$\small{\text{
$
\sin{(\alpha)}=\sqrt{1-\cos^2{(\alpha)}}=\frac{1}{3}\sqrt{5} \qquad
\sin{(\beta)}=\sqrt{1-\cos^2{(\beta)}}= \frac{4}{9}\sqrt{2}
$
}}\\
\small{\text{
$
\sin{(\alpha)} + \sin{(\beta)}= \frac{1}{3}\sqrt{5} +\frac{4}{9}\sqrt{2}
$
}}$$

heureka  Sep 26, 2014
 #4
avatar+238 
+5

 c is not equal to a. because in the equation of x :

a(1-x)^2+2bx+c(1+x^2)=2

a-ax^2+2bx+c+cx^2-2=0

(c-a)x^2+2bx+(a+c-2)=0

when the equation of ax^2+bx+c=0 have 2 ame soloution, it have b^2-4ac=0

in this case, a=c-a b=2b c=a+c-2,so

(2b)^2-4(c-a)(a+c-2)=0

4b^2-4(ac+c^2-2c-a^2-ac+2a)=0

b^2=c^2-2c-a^2+2a

b^2=c^2-2c+1-a^2+2a-1

B^2=(c-1)^2-(a^2-2a+1)

B^2=(c-1)^2-(a-1)^2

so if a=c then (c-1)^2=(a-1)^2, (c-1)^2-(a-1)^2=0, b=0

a side of a triangle is always bigger than 0

so a not equal to c

i also alologize for the question should be a(1-x^2)+2bx+(1+x^2)c=0

i am really sorry about that. I typed it in a wrong way

Good job melody and heureka!

quinn  Sep 26, 2014
 #5
avatar+238 
0

and if a=c    a=2/3b

then sin(a)=(2*square root of 2)/3       cos(a)=1/3.

Good job.it is my bad though.

I am exciting to study and solove problem with you guys.

You guys can understand and use the fomlua very well.That is really smart.

quinn  Sep 26, 2014
 #6
avatar+92806 
+10

thanks Heureka  

I am just going to look at the revised question.

 

$$a(1-x^2)+2bx+(1+x^2)c=0$$

with only one solution

I get to down to 

$$c^2=a^2+b^2$$     

ADDED: (This is the result based on the fact that the descriminant must equal zero)

Which means it is a right angled triangle where c is the hypotenuse.

$$3c=a+3b$$   

$$\\c=\frac{a+3b}{3}\\\\
c^2=\frac{a^2+9b^2+6ab}{9} \qquad (2)\\\\
$Sub (2) into (1)$\\\\
\frac{a^2+9b^2+6ab}{9} =a^2+b^2\\\\
a^2+9b^2+6ab =9a^2+9b^2\\\\
6ab =8a^2\\\\
3b =4a\\\\
b =\frac{4}{3}a\\\\
c=\sqrt{1+\frac{25}{9}}a=\frac{5}{3}a\\\\$$

multiplying all these by 3 and I find that the sides of this triangle are in the ratio 3:4:5

$$sinA+sinB = \dfrac{a+b}{c}=\frac{7}{5}=1.4$$

 

    

Melody  Sep 27, 2014
 #7
avatar+92806 
+10

You have posted 2 questions Quinn, both of them were really good ones that took quite a lot of time.

Both times you made a mistake in the question, hence we 'wasted' our time.  You must be more careful.  

We will lose interest if you continue to do this.

Your questions are great.  If  you can get them to us without error then you will be a very popular member of this forum.  

Melody.

Melody  Sep 27, 2014
 #8
avatar+87333 
+10

Here's my rather (long) take on this one......

So we have

a(1-x^2) + 2bx + (1 + x^2)c  = 0     and    3c = a + 3b  →   a = 3c - 3b   → a = 3(c-b)

So

3(c - b)(1- x^2) + 2bx + (1 +x^2)c = 0     factor the negative from(b-c) and applying it to (1 - x^2) and we have

3(b-c)(x^2 -1) + 2bx + (1+x^2)c = 0  rearranging and factoring we have

[(3(b-c) + c]x^2 + 2bx + [c - 3(b-c)] = 0

(3b -2c)x^2 + 2bx + (4c - 3b) = 0

Now, because this quadratic has a repeated solution, it must mean that its discriminant = 0

So, we have

4b^2 - 4(3b - 2c) (4c - 3b) = 0     divide through by 4

b^2  - (3b - 2c)(4c - 3b) = 0      factor the negative from (3b - 2c)

b^2 + (2c - 3b)(4c - 3b) = 0   

b^2 + 8c^2 - 18bc + 9b^2 = 0    combine like terms

10b^2 -18bc  + 8c^2 = 0  divide through by 2

5b^2 - 9bc + 4c^2  = 0     factor

(5b - 4c) ( b - c) = 0

So,  either.......

b = c  ..... but this is impossible because it makes a = 0   (remember that a = 3(c-b) ....)

or

(5b - 4c) = 0  .......     in which case ......   [ 5b = 4c ]  → [ b = (4/5) c ]

So   a  =  3[c - (4/5)c]  = [3(1/5)c] = (3/5)c

Now, let c  be some convenient length, say, 5 ......  this means that b = 4  and a = 3

And, as Melody has correctly said, we have a classic 3-4-5 "Pythagorean Triple" right triangle !!!

And we could find one angle using the Law of Cosines.....the other angle would then be easily found as well as the sines of A and B .......(This part of the problem isn't particularly challenging !!!)

Note the interesting thing here... the value of the single "solution"  to the quadratic isn't of any import....the fact that the polynomial only has one solution allows us to work with its discriminant to obtain the "answer" to an entirely different problem  !!!

That's my story and I'm sticking to it............

 

  

CPhill  Sep 27, 2014
 #9
avatar+238 
+10

qudratic equation, ax^2+bx+c=0

two different soloution ,b^2-4ac>0

one soloution,b^2-4ac=0

no soloution,b2-4ac<0

quinn  Sep 27, 2014
 #10
avatar+92806 
+10

CPhill has asked me to comment.

------------------------

Yes Chris you have done it exactly the same way as I did. 

I concur with all your conclusions.

It was a very interesting problem.

Thanks for supplying it Quinn.   

Melody  Sep 28, 2014
 #11
avatar+87333 
+10

Just for the heck of it, let's plug the values we found for the triangle's sides back into the original polynomial to see what the solution might be:

(3b -2c)x^2 + 2bx + (4c - 3b) = 0

[3(4) -2(5)] x^2 + 2(4)x  + [4(5) - 3(4)] = 0

2x^2 + 8x + 8 = 0

x^2 + 4x + 4 = 0

(x + 2)  (x + 2) = 0

x + 2 = 0

x = -2

Ah ha !!!....just as quinn noted in his original question.....our polynomial indeed has only one (repeated) root. This seems to further confirm that our answer was correct .......

 

 

CPhill  Sep 28, 2014

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