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# Algebra II Question

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Positive real numbers r, s satisfy the equations r^2+s^2 = 1 and r^4 + s^4 = 7/8. Find rs.

Thanks.

Guest Apr 4, 2018
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### 2+0 Answers

#1
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r^2  + s^2  =  1

r^4  + s^4  = 7/8

s^2  =  1  - r^2

s^4  = (1-r^2)^2   =  r^4 - 2r^2  + 1

So

r^4  +  s^4  = 1

r^4  + r^4 - 2r^2  + 1  = 7/8

2r^4 - 2r^2  + 1/8  = 0            divide through by 2

r^4 - r^2 + 1/16  = 0

r^4 - r^2  = - 1/16      complete the square on r

r^4  - r^2  +1/4  =  -1/16  + 1/4

(r^2 -1/2)^2  =  3/16         take the positive root of both sides

r^2  - 1/2  = √3/4

r^2  =  √3/4  + 1/2

r^2  = [ 2 + √3] / 4      take the positive root

r  =√[ 2 + √3 ] / 2

So.....

s^2  = 1  - r^2

s^2   =  1  - [ 2 + √3] / 4

s^2  = [ 4 - 2 - √3] / 4

s^2  = [ 2 - √3 ] / 4

s  = √[ 2 - √3 ] / 2

So  rs √[ 2 + √3 ] / 2  √[ 2 - √3 ] / 2   =   √ [ (2 + √3) (2 - √3) ]  / 4  = √ [ 4 - 3]  / 4  =   1/4

CPhill  Apr 4, 2018
#2
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Positive real numbers r, s satisfy the equations

r^2+s^2 = 1 and

r^4 + s^4 = 7/8.

Find rs.

$$\begin{array}{|rcll|} \hline (r^2+s^2)^2 &=& r^4+2r^2s^2+s^4 \\ (r^2+s^2)^2 &=& r^4+s^4+2r^2s^2 \\ 2r^2s^2 &=& (r^2+s^2)^2-(r^4+s^4) \quad & | \quad r^2+s^2 = 1 \qquad r^4 + s^4 = \dfrac{7}{8} \\ 2r^2s^2 &=& 1^2-\dfrac{7}{8} \\ 2r^2s^2 &=& \dfrac{1}{8} \\ r^2s^2 &=& \dfrac{1}{16} \\ \mathbf{rs} &\mathbf{=}& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}$$

heureka  Apr 5, 2018

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