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how many roots does the following have

p(x)=(x^2-5)(x^2+4)(x^2+10)(2x+6)

 May 12, 2016

Best Answer 

 #3
avatar+130501 
+5

Actually...there are 7 roots.....3 real and 4 non-real

 

 x = ± √5   and x =  -3       are the real roots

 

And there are 4 non-real roots [ two conjugate pairs, as expected]

 

 

cool cool cool

 May 12, 2016
 #1
avatar+37168 
0

Five

 May 12, 2016
 #2
avatar
+5

Here are the roots:

 

x = -3
x = -sqrt(5)
x = sqrt(5)

Complex roots:
x = -2 i
x = 2 i
x = -i sqrt(10)
x = i sqrt(10)

 May 12, 2016
 #3
avatar+130501 
+5
Best Answer

Actually...there are 7 roots.....3 real and 4 non-real

 

 x = ± √5   and x =  -3       are the real roots

 

And there are 4 non-real roots [ two conjugate pairs, as expected]

 

 

cool cool cool

CPhill May 12, 2016
 #4
avatar+26398 
+5

how many roots does the following have

p(x)=(x^2-5)(x^2+4)(x^2+10)(2x+6)

 

(x25)×(x2+4)×(x2+10)×(2x+6)=(x25)×(x2+4)×(x2+10)×(2(x+3))=(x25)binomial×(x2+4)×(x2+10)×(x+3)×2=(x5)(x+5)×(x2+4)×(x2+10)×(x+3)×2 i2=14i2=4  i2=110i2=10 =(x5)(x+5)×(x24i2)×(x210i2)×(x+3)×2=(x5)(x+5)×(x24i2)binomial×(x210i2)binomial×(x+3)×2=(x5)(x+5)×(x2i)(x+2i)×(x10i)(x+10i)×(x+3)×2=(x5)1. root: 5(x+5)2. root: 5×(x2i)3. root: 2i(x+2i)4. root: 2i×(x10i)5. root: 10i(x+10i)6. root: 10i×(x+3)7. root: 3×2

 

laugh

 May 12, 2016
edited by heureka  May 12, 2016

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