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if r and s are the roots of 2x^2 -5x -7 =0 what is the value of r/s + s/r?

Guest Jul 5, 2014

Best Answer 

 #12
avatar+93305 
+11

Here is another approach.  I suspect that this is how it was meant to be done.

Given the quadratic equation 

$$ax^2+bx+c=0\qquad \mbox{the sum of the roots is } \frac{-b}{a} \mbox{ and the product of the roots is } \frac{c}{a}$$

so

$$2x^2-5x-7=0\qquad r+s= \frac{5}{2}\:\: \mbox{ and }\:\: rs=\frac{-7}{2}$$

Now

$$\dfrac{r}{s}+\dfrac{s}{r}\\\\
=\dfrac{r^2+s^2}{rs}\\\\
=\dfrac{r^2+s^2+2rs-2rs}{rs}\\\\
=\dfrac{(r+s)^2-2rs}{rs}\\\\
=\dfrac{(\frac{5}{2})^2-2\times \frac{-7}{2}}{\frac{-7}{2}}\\\\
=\dfrac{\frac{25}{4}+ \frac{14}{2}}{\frac{-7}{2}}\\\\
=\dfrac{\frac{25}{4}+ \frac{28}{4}}{\frac{-7}{2}}\\\\
=\frac{53}{4}\times\frac{-2}{7}\\\\
=\frac{53}{2}\times\frac{-1}{7}\\\\
=\frac{-53}{14}\\\\
=\:-3\frac{11}{14}$$

Melody  Jul 5, 2014
 #1
avatar+3450 
+5

I'm not quite sure what you mean by "r and s are the roots of..."

Do you mean that r and s are the two values for x in this equation?

This is how I'll solve it. If I misinterpreted the question let me know.

2x2 -5x -7 =0

Let's factor the polynomial and then find the values of x.

(2x+7)(x-1) = 0

------

(2x+7) = 0

2x = -7

x = -3½

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

(x-1) = 0

x = 1

-------

So let's say r = 3½ and s = 1

Let's put this in the equation.

r/s + s/r

3.5/1 + 1/3.5

12.25/3.5 + 1/3.5

13.25/3.5

 

Well...that would be the answer.

Sorry if I totally bombed this, I think the problem was that I didn't really understand the question.

NinjaDevo  Jul 5, 2014
 #2
avatar+4150 
0

YES YOU BOMBED THIS NO POINTS FOR YOU

zegroes  Jul 5, 2014
 #3
avatar+11847 
+3

Zegroes dont be so selfish in letting people take their points!all points are not for u only!

rosala  Jul 5, 2014
 #4
avatar+4150 
0

YES ALL MINE MUHAHAHAHA!!!!!!!!!

zegroes  Jul 5, 2014
 #5
avatar+3450 
0

NinjaDevo  Jul 5, 2014
 #6
avatar+11847 
+3

NOPE!

but u dont have much left to take home as melody has already taken more than half of them!

rosala  Jul 5, 2014
 #7
avatar+4150 
0

MELODYS A DINOSAUR OF COURSE SHE NEEDS TO EAT LIKE 10 TONS OF POINTS (meat) A DAY! 

zegroes  Jul 5, 2014
 #8
avatar+11847 
+3

Nothing to say......!

BTW thank god ND ur bomb isnt real or else it would b**w up all the answers!

rosala  Jul 5, 2014
 #9
avatar+88871 
+5

I think you have the right idea, ND....let's look at the factoring once more...

2x^2 - 5x - 7 = (2x -7) (x + 1)

So the roots are 7/2 and -1

So

(7/2)/(-1) + (-1)/(7/2) =

-7/2 - 2/7   =  -(7/2 + 2/7) = - (49 + 4)/14  = -53/14

I think that might be it ......

 

CPhill  Jul 5, 2014
 #10
avatar+3450 
0

Oh i see, my factoring was off.

Thanks CPhill.

NinjaDevo  Jul 5, 2014
 #11
avatar+88871 
0

I gave you "points" anyway....just a slight mistake....I make 'em all the time!!!  Your approach was correct and that's sometimes more important than just getting the  "right" answer.....

 

 

CPhill  Jul 5, 2014
 #12
avatar+93305 
+11
Best Answer

Here is another approach.  I suspect that this is how it was meant to be done.

Given the quadratic equation 

$$ax^2+bx+c=0\qquad \mbox{the sum of the roots is } \frac{-b}{a} \mbox{ and the product of the roots is } \frac{c}{a}$$

so

$$2x^2-5x-7=0\qquad r+s= \frac{5}{2}\:\: \mbox{ and }\:\: rs=\frac{-7}{2}$$

Now

$$\dfrac{r}{s}+\dfrac{s}{r}\\\\
=\dfrac{r^2+s^2}{rs}\\\\
=\dfrac{r^2+s^2+2rs-2rs}{rs}\\\\
=\dfrac{(r+s)^2-2rs}{rs}\\\\
=\dfrac{(\frac{5}{2})^2-2\times \frac{-7}{2}}{\frac{-7}{2}}\\\\
=\dfrac{\frac{25}{4}+ \frac{14}{2}}{\frac{-7}{2}}\\\\
=\dfrac{\frac{25}{4}+ \frac{28}{4}}{\frac{-7}{2}}\\\\
=\frac{53}{4}\times\frac{-2}{7}\\\\
=\frac{53}{2}\times\frac{-1}{7}\\\\
=\frac{-53}{14}\\\\
=\:-3\frac{11}{14}$$

Melody  Jul 5, 2014
 #13
avatar+20008 
+8

if r and s are the roots of 2x^2 -5x -7 =0 what is the value of r/s + s/r ?

$$ax^2+bx+c=0 \\
\mbox{The roots are: } x_{1,2}= {-b\pm\sqrt{b^2-4ac}\over2a}\\
\mbox{or } x_{1,2}= {-b\pm\sqrt{D}\over2a} \quad \mbox{ set }\quad D=b^2-4ac\\
\mbox{So we have: } r=x_1={-b+\sqrt{D}\over 2a}\\
\mbox{and } s=x_2={-b-\sqrt{D}\over 2a}$$

$$\\ \boxed{{r\over s}+{s\over r} =} {-b+\sqrt{D}\over -b-\sqrt{D}}+{-b-\sqrt{D}\over -b+\sqrt{D}}\\ \\
={(-b+\sqrt{D})^2 +(-b-\sqrt{D})^2\over (-b-\sqrt{D})(-b+\sqrt{D})}\\\\
={2(-b)^2+2(\sqrt{D})^2 \over (-b)^2-(\sqrt{D})^2}\\\\
={2b^2+2D \over b^2-D} \quad | \quad D= b^2-4ac$$

$$\\={2b^2+2b^2-8ac \over b^2-b^2+4ac} \\\\
={4b^2-8ac \over 4ac} \\\\
={b^2 \over ac} - 2 \quad | \quad a=2 \quad b=-5 \quad c=-7$$

$$\\={ (-5)^2 \over 2*(-7) } -2 \\\\
={ 25 \over (-14) } -2 \\\\
={ -25-2*14 \over 14 } \\\\
={ -\frac{53} {14} } \\\\
={ -\frac{14*3+11} {14} } \\\\
={ -3-\frac{11} {14} } \\\\
\boxed{={ -3\frac{11} {14} }}$$

heureka  Jul 7, 2014

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